Algebra 2

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Functions & Their Graphs

📌 Functions & Their Graphs A2 · Composition · Inverse · Domain & range · Piecewise functions

Algebra 2 is really the study of functions as objects you can combine, invert and transform. Fluency with notation, domain and range underpins every later chapter.

● Definition 1.1 — Function

A function assigns to each input exactly one output. We write $y=f(x)$ ; the set of allowed inputs is the domain and the set of resulting outputs is the range.

Composition

(f\circ g)(x)=f\big(g(x)\big)

Inverse

swap

x

and

y

, then solve for

y

Domain rule

no division by 0; no even root of a negative

Inverse check

f\big(f^{-1}(x)\big)=x

Worked example 1.A

Given $f(x)=2x+1$ and $g(x)=x^{2}$ , find $f\big(g(2)\big)$ .

Work inside-out: $g(2)=4$ , then $f(4)=2(4)+1$ :

f\big(g(2)\big)=9.

Worked example 1.B

Find the inverse of $f(x)=2x-6$ .

Write $y=2x-6$ , swap and solve: $x=2y-6 \Rightarrow y=\tfrac{x+6}{2}$ :

f^{-1}(x)=\frac{x+6}{2}.

Transformation of cubic parent

f(x) = x^3

shifted 2 units right and 1 unit up to inflection point

(2, 1)

▲ Algebra 2 Traps & Commandments — Functions & Transformations

Horizontal Shift Intuition: In $f(x - h)$ , subtracting $h$ shifts the graph to the right by $h$ , while $f(x + h)$ shifts to the left. Vertical shifts $f(x) \pm k$ behave normally ( $+k$ is up, $-k$ is down).
Composition Order: $(f \circ g)(x) = f(g(x))$ evaluates $g(x)$ first, then applies $f$ . In general, $f(g(x)) \neq g(f(x))$ . Do not multiply functions when composition is requested!
Inverse Domain/Range Swap: For an invertible function $f$ , the domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ . If the original function is restricted (e.g. $x \ge 0$ ), that restriction dictates the range of $f^{-1}$ .
Even vs. Odd Functions: $f$ is even if $f(-x) = f(x)$ (symmetric across the $y$ -axis, e.g. $x^2, \cos x$ ); $f$ is odd if $f(-x) = -f(x)$ (symmetric about the origin, e.g. $x^3, \sin x$ ).

f(x)=x^2-1

and

g(x)=\sqrt{x+2}

, find (a)

f(g(3))

, (b) the domain of

g\circ f

(a)

g(3)=\sqrt5

;

f(\sqrt5)=5-1=4

. (b)

g(f(x))=\sqrt{x^2-1+2}=\sqrt{x^2+1}

; always real, domain: all reals.

Find the inverse function

f^{-1}(x)

and state its domain and range for

f(x) = \dfrac{3x - 1}{2x + 5}

Write $y = \dfrac{3x - 1}{2x + 5}$ . The domain of $f$ is $x \neq -2.5$ . Its horizontal asymptote is $y = 3/2$ , so the range of $f$ is $y \neq 1.5$ .

Swap $x$ and $y$ and solve for $y$ :

x = \frac{3y - 1}{2y + 5} \implies x(2y + 5) = 3y - 1.

2xy + 5x = 3y - 1 \implies 2xy - 3y = -5x - 1.

y(2x - 3) = -(5x + 1) \implies y = \frac{-5x - 1}{2x - 3} = \frac{5x + 1}{3 - 2x}.

Inverse function: $f^{-1}(x) = \frac{5x + 1}{3 - 2x}.$

Domain of $f^{-1}$ : $x \neq \frac{3}{2}$ (the range of $f$ ).

Range of $f^{-1}$ : $y \neq -\frac{5}{2}$ (the domain of $f$ ).

📝 Chapter Quiz

Questions Level

Quadratics & Complex Numbers

📌 Quadratics & Complex Numbers A2 · Completing the square · Discriminant · Complex roots $a\pm bi$

Every quadratic can be solved — and when the discriminant is negative, the solutions live in the complex numbers. This chapter ties graphing, factoring and the imaginary unit together.

● Definition 2.1 — The imaginary unit

Define $i=\sqrt{-1}$ , so $i^{2}=-1$ . A complex number has the form $a+bi$ with $a,b$ real; add and multiply just as with binomials, replacing $i^{2}$ by $-1$ .

Quadratic formula

x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}

Discriminant

\Delta=b^{2}-4ac

decides the root type

Vertex

x=-\dfrac{b}{2a}

Powers of i

i^{2}=-1,\ i^{3}=-i,\ i^{4}=1

Worked example 2.A

Describe the roots of $x^{2}+2x+5=0$ .

Discriminant $\Delta=2^{2}-4(1)(5)=-16<0$ , so there are no real roots — two complex conjugates. Solving:

x=\frac{-2\pm\sqrt{-16}}{2}=-1\pm 2i.

⊙ Explorer · Discriminant & the parabola's rootsinteractive

b: 2

c: 5

discriminant D-16

root typeNo real roots (complex)

roots-1.00 ± 2.00i

Default matches Worked example 2.A exactly: b = 2, c = 5 gives D = −16, so the graph never touches the x-axis and the roots are the complex pair −1 ± 2i. Drag D up through 0 and the (real) roots appear on the x-axis right when D crosses zero — the discriminant deciding the root type, live.

Worked example 2.B

Multiply $(2+i)(3-i)$ .

6-2i+3i-i^{2}=6+i+1=7+i.

Parabola with vertex at

(2, -1)

, positive discriminant

\Delta = 16 - 12 = 4 > 0

, and distinct real roots at

x = 1

and

x = 3

▲ Algebra 2 Traps & Commandments — Quadratics & Complex Numbers

Powers of i Cycle: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ . For high powers $i^n$ , divide the exponent $n$ by 4 and evaluate using the remainder (e.g. $i^{27} = i^{4(6)+3} = i^3 = -i$ ).
Multiplying Negative Radicals: Never write $\sqrt{-3}\sqrt{-12} = \sqrt{36} = 6$ ! You must extract the imaginary unit first: $(i\sqrt{3})(i\sqrt{12}) = i^2\sqrt{36} = -1(6) = -6$ .
Discriminant Interpretation: For $ax^2 + bx + c = 0$ , $\Delta = b^2 - 4ac$ : if $\Delta > 0$ , two distinct real roots; if $\Delta = 0$ , one repeated real rational root (vertex on $x$ -axis); if $\Delta < 0$ , two non-real complex conjugate roots $a \pm bi$ .
Completing the Square with $a \neq 1$ : Always factor the leading coefficient $a$ out of the quadratic and linear terms before adding $(b/(2a))^2$ . In $2x^2 - 12x$ , factor to $2(x^2 - 6x)$ , then add 9 inside the parentheses (which actually adds $2 \times 9 = 18$ to that side).

Solve

x^2-4x+13=0

and express roots in the form

a\pm bi

Discriminant

=16-52=-36

x=\dfrac{4\pm\sqrt{-36}}{2}=\dfrac{4\pm6i}{2}=2\pm3i

Solve

2x^2 - 6x + 5 = 0

by completing the square, expressing all solutions in exact

a \pm bi

form.

Divide by the leading coefficient 2 or factor it out from the variable terms:

2(x^2 - 3x) = -5 \implies x^2 - 3x = -\frac{5}{2}.

Add $\left(-\frac{3}{2}\right)^2 = \frac{9}{4}$ to both sides:

x^2 - 3x + \frac{9}{4} = -\frac{5}{2} + \frac{9}{4} = -\frac{10}{4} + \frac{9}{4} = -\frac{1}{4}.

Write as a perfect square and take the square root of both sides:

\left(x - \frac{3}{2}\right)^2 = -\frac{1}{4} \implies x - \frac{3}{2} = \pm\sqrt{-\frac{1}{4}} = \pm\frac{i}{2}.

x = \frac{3}{2} \pm \frac{1}{2}i.

📝 Chapter Quiz

Questions Level

Polynomial Functions

📌 Polynomial Functions A2 · Factor theorem · Remainder theorem · Rational roots · End behaviour

Beyond quadratics lie cubics, quartics and higher. Two questions dominate: where are the zeros, and what does the graph do at its ends? The leading term answers the second on its own.

▲ Reference — End behaviour & the factor theorem

The end behaviour of a polynomial is governed entirely by its leading term $a x^{n}$ : the sign of $a$ and the parity of $n$ . And $(x-r)$ is a factor exactly when $r$ is a zero ( $f(r)=0$ ).

Change the degree and the leading coefficient below. Watch how the arms of the graph flip with the sign of $a$ , and how an even degree sends both arms the same way while an odd degree sends them opposite ways.

⊙ Explorer · Degree, sign & end behaviourinteractive

Degree

n

Leading coefficient a = 1

as x → +∞—

as x → −∞—

turning points—

Remainder theorem

dividing by

(x-r)

leaves remainder

f(r)

Factor theorem

f(r)=0 \iff (x-r)

is a factor

Zeros

a degree-

n

polynomial has

n

zeros (with multiplicity)

Worked example 3.A

Find the zeros of $f(x)=x^{3}-9x$ .

Factor out $x$ , then a difference of squares:

x(x^{2}-9)=x(x-3)(x+3)=0 \ \Rightarrow\ x=0,\ 3,\ -3.

Worked example 3.B

Find the remainder when $x^{2}+3x-4$ is divided by $(x-1)$ .

By the remainder theorem, evaluate at $x=1$ :

1+3-4=0.

Cubic polynomial with a single root at

x = -1

(odd multiplicity 1, crosses axis) and a double root at

x = 2

(even multiplicity 2, touches and turns).

▲ Algebra 2 Traps & Commandments — Polynomials & Roots

Remainder Theorem Evaluation Sign: When dividing by $(x - c)$ , the remainder is $f(c)$ . Notice the sign flip: dividing by $(x + 3)$ requires evaluating $f(-3)$ , not $f(3)$ .
Root Multiplicity and Axis Crossing: A root with odd multiplicity (e.g. 1, 3) crosses the $x$ -axis; a root with even multiplicity (e.g. 2, 4) is tangent to the axis (touches and rebounds without crossing).
Missing Degree Terms in Long/Synthetic Division: Always check that exponents decrease consecutively: $x^4, x^3, x^2, x^1, x^0$ . If a term is missing (e.g. $x^3 - 7x + 6$ ), you must insert a $0x^2$ placeholder in the division tableau.
Rational Root Theorem List: All possible rational zeros must have the form $\pm p/q$ , where $p$ divides the constant term $a_0$ and $q$ divides the leading coefficient $a_n$ . Never invert this ratio!

Show that

(x-2)

is a factor of

p(x)=x^3-3x^2+4

, then fully factorise

p(x)

p(2)=8-12+4=0

✓. Divide:

p(x)=(x-2)(x^2-x-2)=(x-2)(x-2)(x+1)=(x-2)^2(x+1)

Find all real and complex roots of the polynomial

P(x) = 2x^3 + 3x^2 - 8x + 3

By the Rational Root Theorem, possible rational roots are $\pm\{1, 3\}/\{1, 2\} = \pm\{1, 3, 1/2, 3/2\}$ .

Test $x = 1$ : $P(1) = 2(1)^3 + 3(1)^2 - 8(1) + 3 = 2 + 3 - 8 + 3 = 0. \quad \checkmark$

So $(x - 1)$ is a factor. Divide $P(x)$ by $(x - 1)$ using synthetic division:

2x^3 + 3x^2 - 8x + 3 = (x - 1)(2x^2 + 5x - 3).

Factor the remaining quadratic:

2x^2 + 5x - 3 = (2x - 1)(x + 3).

Setting each factor to zero yields the complete set of roots: $x = 1, \quad x = \frac{1}{2}, \quad x = -3.$

📝 Chapter Quiz

Questions Level

Rational Expressions

📌 Rational Expressions A2 · Asymptotes · Holes · Graphing rational functions · Partial fractions

A rational function is a ratio of polynomials. The new feature is the asymptote — a line the graph approaches but never meets — produced by values that make the denominator zero.

● Definition 4.1 — Asymptotes

For $y=\dfrac{1}{x-h}+k$ : a vertical asymptote at $x=h$ (denominator zero) and a horizontal asymptote at $y=k$ (the end-behaviour value).

Slide the asymptotes around. The two amber dashed lines are $x=h$ and $y=k$ ; the curve hugs them in every corner.

⊙ Explorer · Rational function & asymptotesinteractive

Vertical shift of asymptote h

Horizontal asymptote k

vertical asymptote—

horizontal asymptote—

y = 1 / (x − 1)

Worked example 4.A

Simplify $\dfrac{x^{2}-9}{x-3}$ .

Factor and cancel (for $x\neq 3$ ):

\frac{(x-3)(x+3)}{x-3}=x+3.

Worked example 4.B

Solve $\dfrac{1}{x}+\dfrac{1}{2}=1$ .

Then $\dfrac{1}{x}=\dfrac12$ , so:

x=2.

▲ Rational expressions and equations

A rational expression is a ratio of polynomials. Simplify by factoring and cancelling common factors. Multiply LCD to solve rational equations; check for extraneous solutions.

Simplify

Factor numerator and denominator; cancel common factors

Add/subtract

Find LCD; rewrite each fraction; combine numerators

Multiply

Multiply numerators and denominators; simplify

Solve equation

Multiply both sides by LCD; solve; check for excluded values

Worked example

Simplify $\dfrac{x^2-4}{x^2-x-2}$ .

\frac{(x-2)(x+2)}{(x-2)(x+1)}=\frac{x+2}{x+1},\quad x\neq 2,-1.

Hyperbolic branches of

y = \frac{2x+1}{x-3}

with vertical asymptote at

x = 3

and horizontal asymptote at

y = 2

▲ Algebra 2 Traps & Commandments — Rational Expressions & Equations

Extraneous Solutions from Clearing Denominators: Multiplying an equation by the LCD clears fractions but may produce solutions that make an original denominator zero. Any solution making a denominator zero is extraneous and must be discarded.
Illegal Term Cancellation: Never cancel individual addends (e.g. canceling $x$ in $\frac{x + 2}{x + 5}$ ). You may only cancel complete factors after factoring the entire numerator and denominator into products.
Removable Discontinuity (Hole) vs. Vertical Asymptote: If a factor $(x - c)$ cancels completely from both numerator and denominator, $x = c$ is a hole at $(c, g(c))$ , not a vertical asymptote. Only uncanceled denominator zeros form vertical asymptotes.
Dividing Compound Fractions: To divide fractions, multiply by the reciprocal of the divisor: $\frac{A/B}{C/D} = \frac{A}{B} \cdot \frac{D}{C}$ . Factor all four polynomials before multiplying to simplify cancellations.

Sketch

y=\dfrac{2x+1}{x-3}

, showing all asymptotes and intercepts.

Vertical asymptote:

x=3

. Horizontal asymptote:

y=2

(leading coefficients).

y

-intercept:

y=-1/3

x

-intercept:

x=-1/2

Solve the rational equation for

x

\frac{x}{x - 2} + \frac{1}{x + 3} = \frac{5}{x^2 + x - 6}.

Check for extraneous solutions.

Factor the quadratic denominator: $x^2 + x - 6 = (x - 2)(x + 3)$ . The LCD is $(x - 2)(x + 3)$ , with restricted domain $x \neq 2, -3$ .

Multiply both sides by $(x - 2)(x + 3)$ :

x(x + 3) + 1(x - 2) = 5.

x^2 + 3x + x - 2 = 5 \implies x^2 + 4x - 7 = 0.

Apply the quadratic formula:

x = \frac{-4 \pm \sqrt{16 - 4(1)(-7)}}{2} = \frac{-4 \pm \sqrt{16 + 28}}{2} = \frac{-4 \pm \sqrt{44}}{2} = \frac{-4 \pm 2\sqrt{11}}{2} = -2 \pm \sqrt{11}.

Check restrictions: $-2 \pm \sqrt{11} \approx 1.317$ and $-5.317$ . Neither equals 2 or -3, so both solutions are valid: $x = -2 + \sqrt{11} \quad \text{and} \quad x = -2 - \sqrt{11}.$

📝 Chapter Quiz

Questions Level

Exponential & Logarithmic Functions

📌 Exponential & Logarithmic Functions A2 · Laws of logs · Solving $e^x=k$ · Exponential growth/decay

Exponential functions model growth and decay; logarithms are their inverses, and the tool that lets you solve for an exponent. The log laws turn products into sums.

● Definition 5.1 — Logarithm

$\log_{a} b=c$ means exactly $a^{c}=b$ (with $a>0,\ a\neq 1$ ). So a logarithm is an exponent.

⊙ Explorer · Exponential growth & decayinteractive

rate k: 0.5

behaviorgrowth

y at x = 11.649

y = e^(kx): positive k gives growth, negative k gives decay, k = 0 is constant. Every exponential function in this chapter — including 2^x from Worked example 5.A — has this same growth/decay shape once rewritten in base e.

Product

\log_a(xy)=\log_a x+\log_a y

Quotient

\log_a\!\tfrac{x}{y}=\log_a x-\log_a y

Power

\log_a(x^{k})=k\log_a x

Change of base

\log_a x=\dfrac{\ln x}{\ln a}

Worked example 5.A

Solve $2^{x}=16$ .

Write 16 as a power of 2: $2^{x}=2^{4}$ :

x=4.

Worked example 5.B

Condense $2\log x+\log y$ to a single logarithm.

Power rule then product rule:

2\log x+\log y=\log\!\big(x^{2}y\big).

Exponential growth with base

b = 2 > 1

contrasted with exponential decay with base

b = 0.5 < 1

, both sharing

y

-intercept

(0, 1)

and horizontal asymptote

y = 0

▲ Algebra 2 Traps & Commandments — Exponentials & Logarithms

Logarithmic Domain Restriction: The argument of any logarithm must be strictly positive ( $x > 0$ ). When solving logarithmic equations by condensing (e.g. $\log_2(x) + \log_2(x - 3) = 2$ ), always verify candidate roots in the original uncombined equation to detect extraneous negative or zero arguments.
Common False Logarithmic Laws: Remember that $\log(A + B) \neq \log A + \log B$ and $\frac{\log A}{\log B} \neq \log(A - B)$ . The product property is $\log(AB) = \log A + \log B$ , and the quotient property is $\log(A/B) = \log A - \log B$ .
Quadratic Forms with Exponentials: Equations like $4^x - 5(2^x) + 4 = 0$ are quadratic in disguise. Substitute $u = 2^x$ (so $4^x = (2^x)^2 = u^2$ ) to solve $u^2 - 5u + 4 = 0$ , then solve $2^x = u$ . Reject any $u \le 0$ .
Change of Base Formula: $\log_b a = \frac{\ln a}{\ln b} = \frac{\log a}{\log b}$ . The old base goes in the denominator, and the argument goes in the numerator.

Solve

2^{x+1}=5^{x-1}

, giving your answer to 3 significant figures.

Take logs:

(x+1)\ln2=(x-1)\ln5

x\ln2-x\ln5=-\ln5-\ln2

x=\dfrac{-\ln10}{\ln2-\ln5}=\dfrac{-\ln10}{\ln(2/5)}\approx2.48

Solve the exponential equation

4^{x+1} - 5(2^x) + 1 = 0

for all real values of

x

Rewrite the first term in terms of base 2: $4^{x+1} = 4^1 \cdot (2^2)^x = 4 \cdot (2^x)^2.$

Substitute $u = 2^x$ (with constraint $u > 0$ ):

4u^2 - 5u + 1 = 0.

Factor the quadratic:

(4u - 1)(u - 1) = 0 \implies u = \frac{1}{4} \quad \text{or} \quad u = 1.

Back-substitute $2^x = u$ :

$2^x = \frac{1}{4} = 2^{-2} \implies x = -2$
$2^x = 1 = 2^0 \implies x = 0$

Both solutions are real and valid: $x = -2 \quad \text{and} \quad x = 0.$

📝 Chapter Quiz

Questions Level

Radicals & Rational Exponents

📌 Radicals & Rational Exponents A2 · Simplifying radicals · Rationalising · Solving radical equations

Roots and fractional exponents are two notations for the same idea. Converting between them turns awkward radical problems into routine exponent arithmetic.

● Definition 6.1 — Rational exponent

$a^{m/n}=\sqrt[n]{a^{m}}=\big(\sqrt[n]{a}\big)^{m}$ . The denominator is the root; the numerator is the power.

Product

\sqrt{a}\,\sqrt{b}=\sqrt{ab}

Quotient

\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}

Rationalising

multiply by

\dfrac{\sqrt{b}}{\sqrt{b}}

Worked example 6.A

Evaluate $8^{2/3}$ .

Take the cube root first, then square:

\big(\sqrt[3]{8}\big)^{2}=2^{2}=4.

Worked example 6.B

Solve $\sqrt{x+3}=5$ .

Square both sides: $x+3=25$ :

x=22.

Square root function

y = \sqrt{x+3} - 1

starting at endpoint

(-3, -1)

with natural domain

x \ge -3

and range

y \ge -1

▲ Algebra 2 Traps & Commandments — Radicals & Rational Exponents

Extraneous Roots from Squaring: Squaring both sides of an equation introduces spurious solutions (e.g. $A = B \implies A^2 = B^2$ , but $(-3)^2 = 3^2$ does not mean $-3 = 3$ ). You must plug every candidate back into the original un-squared radical equation!
Principal Non-Negative Root: $\sqrt{x}$ always denotes the non-negative square root. If an equation simplifies to $\sqrt{x} = -k$ (where $k > 0$ ), there are immediately no real solutions.
Absolute Value on Even Roots: $\sqrt{x^2} = |x|$ , not simply $x$ . In general, $\sqrt[n]{x^n} = |x|$ whenever $n$ is an even integer.
Isolate the Radical First: Never square an equation while a radical is added to other terms (e.g. $\sqrt{x+1} + 3 = x$ ). Always isolate the radical as $\sqrt{x+1} = x - 3$ before squaring.

Solve

\sqrt{3x+1}=x-1

, checking for extraneous solutions.

Square both sides:

3x+1=(x-1)^2=x^2-2x+1\Rightarrow x^2-5x=0\Rightarrow x=0

x=5

. Check

x=0

1=-1

✗ (extraneous). Check

x=5

4=4

✓. Solution:

x=5

Solve the radical equation

\sqrt{2x + 7} - x = 2

for all real solutions and confirm extraneous roots.

Isolate the radical term:

\sqrt{2x + 7} = x + 2.

Square both sides:

2x + 7 = (x + 2)^2 = x^2 + 4x + 4.

x^2 + 2x - 3 = 0 \implies (x + 3)(x - 1) = 0 \implies x = -3 \quad \text{or} \quad x = 1.

Test in the original un-squared equation:

Check $x = -3$ : $\sqrt{2(-3) + 7} - (-3) = \sqrt{1} + 3 = 1 + 3 = 4 \neq 2. \quad \text{(Extraneous ✗)}$
Check $x = 1$ : $\sqrt{2(1) + 7} - 1 = \sqrt{9} - 1 = 3 - 1 = 2. \quad \text{(Valid ✓)}$

The only valid real solution is $x = 1$ .

📝 Chapter Quiz

Questions Level

Sequences & Series

📌 Sequences & Series A2 · Arithmetic & geometric · Sigma notation · Binomial theorem

Arithmetic sequences add a constant; geometric sequences multiply by one. Summing their terms gives series — including the surprising fact that an infinite geometric series can have a finite total.

● Definition 7.1 — Arithmetic vs geometric

An arithmetic sequence has a common difference $d$ ; a geometric sequence has a common ratio $r$ . Their $n$ th terms are $a_1+(n-1)d$ and $a_1 r^{\,n-1}$ .

Arithmetic sum

S_n=\tfrac{n}{2}\big(a_1+a_n\big)

Geometric sum

S_n=a_1\dfrac{1-r^{n}}{1-r}

Infinite geometric

S=\dfrac{a_1}{1-r}\ \ (|r|<1)

Worked example 7.A

Find the 10th term of the arithmetic sequence with $a_1=2$ , $d=3$ .

a_{10}=2+(10-1)(3)=29.

Worked example 7.B

Find the sum of $8+4+2+1+\cdots$

Geometric with $a_1=8$ , $r=\tfrac12$ :

S=\frac{8}{1-\tfrac12}=16.

Discrete terms of arithmetic sequence

a_n = 2 + (n - 1)(3) = 3n - 1

lying along a line with constant slope equal to common difference

d = 3

▲ Algebra 2 Traps & Commandments — Sequences, Series & Binomial Theorem

Arithmetic vs. Geometric Recognition: Arithmetic sequences possess a constant difference $d = a_n - a_{n-1}$ (linear growth); geometric sequences possess a constant ratio $r = a_n / a_{n-1}$ (exponential growth). Never confuse the summation formulas!
Counting Terms in Sigma Notation: The number of terms in $\sum_{k=m}^n a_k$ is $n - m + 1$ , not $n - m$ . For instance, $\sum_{k=3}^8 a_k$ contains $8 - 3 + 1 = 6$ terms.
Infinite Sum Existence Condition: The infinite geometric series formula $S = \frac{a_1}{1 - r}$ converges if and only if $|r| < 1$ . If $|r| \ge 1$ , the series diverges and has no finite sum.
Binomial Coefficient Sign Alternation: In $(a - b)^n$ , the negative sign is bound to $b$ : term $k$ is $\binom{n}{k} a^{n-k} (-b)^k = (-1)^k \binom{n}{k} a^{n-k} b^k$ . Terms alternate signs, starting positive for $k = 0$ .

Find the coefficient of

x^2

in the expansion of

(3-2x)^5

\binom{5}{2}(3)^3(-2x)^2=10\times27\times4x^2=1080x^2

. Coefficient:

1080

In an arithmetic sequence, the 4th term is 17 and the 10th term is 47. Find the first term

a_1

, the common difference

d

, and the sum of the first 25 terms

S_{25}

Using $a_n = a_1 + (n - 1)d$ :

a_4 = a_1 + 3d = 17,

a_{10} = a_1 + 9d = 47.

Subtract the first equation from the second:

6d = 30 \implies d = 5.

Substitute $d = 5$ into the first equation:

a_1 + 3(5) = 17 \implies a_1 = 17 - 15 = 2.

Find the sum of the first 25 terms using $S_n = \frac{n}{2}[2a_1 + (n - 1)d]$ :

S_{25} = \frac{25}{2}[2(2) + 24(5)] = \frac{25}{2}[4 + 120] = \frac{25}{2}(124) = 25 \times 62 = 1550.

📝 Chapter Quiz

Questions Level

Conic Sections

📌 Conic Sections A2 · Parabola · Ellipse · Hyperbola · Standard forms & transformations

Slice a cone and the cross-section is a circle, ellipse, parabola or hyperbola. Each has a standard equation you can read a graph straight off of.

▲ Reference — The four conics

Circle $x^{2}+y^{2}=r^{2}$ ; ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ ; parabola $y=\dfrac{x^{2}}{4p}$ ; hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ . A plus sign closes the curve; a minus sign opens it.

Switch between the four and stretch the defining parameter. Notice how the ellipse becomes a circle when its two radii match, and how the hyperbola's minus sign splits it into two branches.

⊙ Explorer · The four conic sectionsinteractive

Conic

Shape parameter

Equation: x²/16 + y²/9 = 1

Worked example 8.A

Give the centre and radius of $(x-2)^{2}+(y+3)^{2}=16$ .

Standard circle form $(x-h)^{2}+(y-k)^{2}=r^{2}$ :

\text{centre } (2,-3),\qquad r=4.

Worked example 8.B

Find the vertices of $\dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1$ .

The $x^{2}$ term is positive, so the branches open left/right with $a=4$ :

(\pm 4,\,0).

▲ Conic sections

Conics are curves formed by intersecting a plane with a double cone: circles, ellipses, parabolas, and hyperbolas.

Circle

(x-h)^2+(y-k)^2=r^2

, centre

(h,k)

, radius

r

Ellipse

\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1

Parabola

y=a(x-h)^2+k

x=a(y-k)^2+h

Hyperbola

\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1

; asymptotes

y=\pm\dfrac{b}{a}x

Worked example

Identify $4x^2+9y^2=36$ and find its semi-axes.

\frac{x^2}{9}+\frac{y^2}{4}=1 \implies a=3,\;b=2.\text{ Ellipse.}

Circle centered at

(2, -1)

with radius

r = 3

, given by standard form

(x - 2)^2 + (y + 1)^2 = 9

▲ Algebra 2 Traps & Commandments — Conic Sections

Completing Squares on Circles & Ellipses: When completing squares on $Ax^2 + Cy^2 + Dx + Ey + F = 0$ , factor out coefficients before adding half-squared values, and multiply by the factored coefficient when balancing the right-hand side.
Focal Distances Sign Rule: For an ellipse, $c^2 = a^2 - b^2$ (foci are inside, so $c < a$ ). For a hyperbola, $c^2 = a^2 + b^2$ (foci are outside, so $c > a$ ).
Major Axis Orientation in Ellipses: $a^2$ is always the larger denominator. If the larger denominator is under $x^2$ , the major axis is horizontal; if under $y^2$ , the major axis is vertical.
Tangent Lines to Circles: A line tangent to a circle at point $P(x_1, y_1)$ is perpendicular to the radius connecting center $C$ to $P$ . The slope of the tangent line is the negative reciprocal of the radius slope $m_{\text{rad}}$ .

Write

9x^2+4y^2-36x+8y+4=0

in standard form and identify the conic.

Complete the square:

9(x-2)^2+4(y+1)^2=36\Rightarrow\dfrac{(x-2)^2}{4}+\dfrac{(y+1)^2}{9}=1

. Ellipse, centre

(2,-1)

a=3

b=2

Write the general circle equation

x^2 + y^2 - 8x + 6y + 9 = 0

in standard form, identify its center and radius, and find the equation of the tangent line to the circle at the point

P(4, 1)

Complete the square for $x$ and $y$ :

(x^2 - 8x + 16) + (y^2 + 6y + 9) = -9 + 16 + 9 = 16.

(x - 4)^2 + (y + 3)^2 = 16.

Center is $C(4, -3)$ and radius is $r = \sqrt{16} = 4$ .

Check the point $P(4, 1)$ : $(4 - 4)^2 + (1 + 3)^2 = 0 + 16 = 16$ $\checkmark$ .

The radius connects $C(4, -3)$ to $P(4, 1)$ , which is a vertical line $x = 4$ . Therefore, the tangent line through $P(4, 1)$ is horizontal: $y = 1.$

📝 Chapter Quiz

Questions Level

Trigonometric Functions

📌 Trigonometric Functions A2 · Unit circle · Graphs of $\sin/\cos/\tan$ · Inverse trig · Identities

The unit circle extends sine, cosine and tangent to all angles and produces periodic graphs.

▲ Reference — Unit circle & graphs

On the unit circle a point is $(\cos\theta,\sin\theta)$ . Radians: $180^{\circ}=\pi$ . The graphs of $\sin x$ and $\cos x$ have period $2\pi$ ; $\tan x$ has period $\pi$ .

⊙ Explorer · The unit circleinteractive

angle θ: 60

cos θ0.500

sin θ0.866

θ in radians1.047

A point on the unit circle is (cos θ, sin θ) — the Reference box's definition, live. Drag θ to 90° and compare the radian readout to Worked example 9.A's π/2.

Unit circle

(\cos\theta,\sin\theta)

Radians

180^{\circ}=\pi

Identity

\sin^{2}\theta+\cos^{2}\theta=1

Period

2\pi

(sin, cos)

Worked example 9.A

Convert $90^{\circ}$ to radians.

\frac{\pi}{2}.

Worked example — Proving a trig identity

Prove that $\dfrac{\sin2\theta}{1+\cos2\theta}=\tan\theta$ .

\text{LHS}=\frac{2\sin\theta\cos\theta}{1+(2\cos^2\theta-1)}=\frac{2\sin\theta\cos\theta}{2\cos^2\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta=\text{RHS}.

Unit circle angle

\theta = 210°

in Quadrant III with reference angle

30°

, showing coordinates

(-\sqrt{3}/2, -1/2)

▲ Algebra 2 Traps & Commandments — Trigonometry & Periodic Functions

Radian vs. Degree Conversions: Radians = Degrees $\times \frac{\pi}{180°}$ ; Degrees = Radians $\times \frac{180°}{\pi}$ . Ensure your calculator is set to the correct mode for each calculation.
Quadrant Signs (ASTC): In Q1 (All positive), Q2 (Sine positive), Q3 (Tangent positive), Q4 (Cosine positive). A negative $\sin\theta$ must place $\theta$ in Q3 or Q4.
Amplitude vs. Vertical Stretch: For $y = A\sin(Bx)$ , the amplitude is strictly non-negative: $|A|$ . A negative coefficient (e.g. $-3\sin x$ ) indicates a vertical reflection, but the amplitude remains $3$ .
Period of Tangent vs. Sine/Cosine: The period of $\sin(Bx)$ and $\cos(Bx)$ is $\frac{2\pi}{|B|}$ . However, the period of $\tan(Bx)$ is $\frac{\pi}{|B|}$ . Do not use $2\pi$ for tangent!

Find all solutions of

2\cos^2\theta+\cos\theta-1=0

for

0°\leq\theta\leq360°

Factor:

(2\cos\theta-1)(\cos\theta+1)=0\Rightarrow\cos\theta=\tfrac{1}{2}

\cos\theta=-1

. Solutions:

\theta=60°,300°,180°

\sin\theta = -\frac{4}{5}

and

\theta

lies in Quadrant III, find the exact values of

\cos\theta

\tan\theta

, and

\sin(2\theta)

Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ :

\left(-\frac{4}{5}\right)^2 + \cos^2\theta = 1 \implies \frac{16}{25} + \cos^2\theta = 1 \implies \cos^2\theta = \frac{9}{25}.

Since $\theta$ is in Quadrant III, cosine is negative: $\cos\theta = -\frac{3}{5}.$

Tangent is positive in Quadrant III: $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-4/5}{-3/5} = \frac{4}{3}.$

Double-angle formula for sine: $\sin(2\theta) = 2\sin\theta\cos\theta = 2\left(-\frac{4}{5}\right)\left(-\frac{3}{5}\right) = \frac{24}{25}.$

📝 Chapter Quiz

Questions Level

Matrices & Systems

📌 Matrices & Systems A2 · Matrix operations · Determinant · Inverse · Solving systems

A matrix is a rectangular array of numbers; matrices add, multiply, and solve linear systems.

▲ Reference — Matrix basics

Add matrices of the same size entrywise. For a $2\times2$ matrix with rows $(a,b)$ and $(c,d)$ , the determinant is $ad-bc$ . A system $A\mathbf{x}=\mathbf{b}$ solves as $\mathbf{x}=A^{-1}\mathbf{b}$ when the determinant is non-zero.

Order

rows × columns

Add

entrywise

Determinant

ad-bc

Solve

\mathbf{x}=A^{-1}\mathbf{b}

Worked example 10.A

Find the determinant of the matrix with rows $(2,1)$ and $(3,4)$ .

2(4)-1(3)=5.

Worked example — Solving a 3×3 system

Use the inverse matrix method to solve $2x+y=5$ , $x-3y=1$ .

A=\begin{pmatrix}2&1\\1&-3\end{pmatrix},\;\det A=-6-1=-7,\;A^{-1}=\frac{1}{-7}\begin{pmatrix}-3&-1\\-1&2\end{pmatrix}.

\begin{pmatrix}x\\y\end{pmatrix}=A^{-1}\begin{pmatrix}5\\1\end{pmatrix}=\frac{1}{-7}\begin{pmatrix}-16\\-3\end{pmatrix}=\begin{pmatrix}16/7\\3/7\end{pmatrix}.

Geometric representation of the

2 \times 2

linear system, intersecting at the unique point

(x, y) = (16/7, 3/7)

because coefficient determinant

\det(A) = -7 \neq 0

▲ Algebra 2 Traps & Commandments — Matrices & Systems

Matrix Multiplication Non-Commutativity: In general, matrix multiplication is not commutative: $AB \neq BA$ . Always maintain the exact order of factors. Furthermore, the product $AB$ is defined only when the number of columns in $A$ equals the number of rows in $B$ .
Singular Matrices & Inverses: A square matrix $A$ has an inverse $A^{-1}$ if and only if its determinant is non-zero ( $\det A \neq 0$ ). If $\det A = 0$ , the matrix is singular and the corresponding linear system either has no solution (parallel lines) or infinitely many solutions (coincident lines).
2×2 Inverse Formula Signs: For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , $A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ . Swap the main diagonal entries ( $a$ and $d$ ), and negate the off-diagonal entries ( $-b$ and $-c$ ).
Cramer's Rule Denominator: In Cramer's Rule, $x = \frac{\det(A_x)}{\det(A)}$ and $y = \frac{\det(A_y)}{\det(A)}$ . The main determinant $\det(A)$ is always the denominator. If $\det(A) = 0$ , Cramer's Rule cannot be used.

Find the value(s) of

k

for which

A=\begin{pmatrix}k&2\\3&k\end{pmatrix}

is singular.

\det A=k^2-6=0\Rightarrow k=\pm\sqrt6

Solve the linear system using matrix inversion or Cramer's Rule:

3x - 2y = 8

5x + 4y = 6

Write the system in matrix form $A\mathbf{x} = \mathbf{b}$ :

\begin{pmatrix} 3 & -2 \\ 5 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ 6 \end{pmatrix}.

Calculate the determinant of coefficient matrix $A$ :

\det A = (3)(4) - (-2)(5) = 12 - (-10) = 22.

Find the inverse matrix $A^{-1}$ :

A^{-1} = \frac{1}{22} \begin{pmatrix} 4 & 2 \\ -5 & 3 \end{pmatrix}.

Multiply $A^{-1} \mathbf{b}$ :

\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{22} \begin{pmatrix} 4(8) + 2(6) \\ -5(8) + 3(6) \end{pmatrix} = \frac{1}{22} \begin{pmatrix} 32 + 12 \\ -40 + 18 \end{pmatrix} = \frac{1}{22} \begin{pmatrix} 44 \\ -22 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}.

Solution: $x = 2$ and $y = -1$ .

📝 Chapter Quiz

Questions Level

Probability & Statistics

📌 Probability & Statistics A2 · Counting · Probability rules · Normal distribution · Data analysis

Counting principles and probability describe and predict uncertain outcomes.

▲ Reference — Counting & probability

Permutations ${}_nP_r=\tfrac{n!}{(n-r)!}$ ; combinations ${}_nC_r=\tfrac{n!}{r!\,(n-r)!}$ . Probability $=\tfrac{\text{favourable}}{\text{total}}$ .

Factorial

n!

Permutation

\tfrac{n!}{(n-r)!}

Combination

\tfrac{n!}{r!(n-r)!}

Probability

\tfrac{\text{fav}}{\text{total}}

Worked example 11.A

Compute ${}_5C_2$ .

\frac{5!}{2!\,3!}=10.

Worked example — Normal distribution application

Heights are normally distributed with $\mu=170$ cm and $\sigma=10$ cm. Find the probability that a randomly chosen person has height between 160 and 185 cm.

P(160\leq X\leq185)=P(-1\leq Z\leq1.5)=\Phi(1.5)-\Phi(-1)=0.9332-0.1587=0.7745.

Normal distribution model with mean

\mu = 170

and standard deviation

\sigma = 10

, showing shaded area

P(160 \le X \le 185) \approx 0.7745

(

-1 \le Z \le 1.5

▲ Algebra 2 Traps & Commandments — Probability & Statistics

Permutations vs. Combinations: Use permutations ${}_nP_r = \frac{n!}{(n-r)!}$ when order or position matters (e.g. race finishes, assigned officer roles, passwords). Use combinations ${}_nC_r = \frac{n!}{r!(n-r)!}$ when order is irrelevant (e.g. groups, hands of cards, committee selections).
Mutually Exclusive vs. Independent: Events $A$ and $B$ are mutually exclusive if they cannot happen together: $P(A \cap B) = 0$ , giving $P(A \cup B) = P(A) + P(B)$ . Events are independent if one event's outcome has no effect on the other: $P(A \cap B) = P(A) \cdot P(B)$ . Do not conflate the two concepts!
Empirical Rule Benchmarks: For any normal distribution: $\approx 68.3\%$ falls within $\mu \pm 1\sigma$ , $\approx 95.4\%$ falls within $\mu \pm 2\sigma$ , and $\approx 99.7\%$ falls within $\mu \pm 3\sigma$ .
Z-Score Standardizing: Always compute $z = \frac{x - \mu}{\sigma}$ . A negative $z$ -score signifies an observation below the mean; never omit the negative sign.

In a class of 200 students taking a test with mean 65 and SD 12 (normally distributed), how many students would you expect to score above 80?

P(X>80)=P(Z>1.25)=1-0.8944=0.1056

. Expected:

200\times0.1056\approx21

students.

A committee of 4 members is to be selected at random from a club of 7 seniors and 5 juniors. (a) What is the total number of possible committees? (b) Find the probability that the committee contains exactly 2 seniors and 2 juniors.

(a) Order does not matter, so the total number of ways to choose 4 members from 12 total people is: $\text{Total} = \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495.$

(b) Number of ways to choose 2 seniors from 7: $\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21.$

Number of ways to choose 2 juniors from 5: $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10.$

By the fundamental counting principle, favorable outcomes: $21 \times 10 = 210.$

Probability: $P(\text{2 seniors, 2 juniors}) = \frac{210}{495} = \frac{14}{33} \approx 0.4242 \quad (42.4\%).$

📝 Chapter Quiz

Questions Level

Practice set

Fifty mixed problems spanning all eight chapters — functions through conic sections — and tagged by difficulty. Click any problem to reveal a full worked solution.

If $f(x)=2x+3$ , find $f(4)$ .

Basic

2(4)+3=11.

If $f(x)=x^{2}-1$ , find $f(-3)$ .

Basic

(-3)^{2}-1=8.

State the domain of $f(x)=\sqrt{x-2}$ .

Intermediate

The radicand must be $\ge 0$ :

x\ge 2.

If $f(x)=2x+1$ and $g(x)=x^{2}$ , find $f\big(g(2)\big)$ .

Intermediate

$g(2)=4$ , then $f(4)=9$ :

9.

Find the inverse of $f(x)=2x-6$ .

Intermediate

f^{-1}(x)=\frac{x+6}{2}.

If $f(x)=x^{2}$ and $g(x)=x+3$ , find $(g\circ f)(x)$ .

Advanced

$g\big(f(x)\big)=g\big(x^{2}\big)=x^{2}+3$ :

x^{2}+3.

Solve $x^{2}=49$ .

Basic

x=\pm 7.

Solve $x^{2}+6x+5=0$ .

Intermediate

$(x+1)(x+5)=0$ :

x=-1\ \text{or}\ x=-5.

Find the vertex of $y=x^{2}-4x+1$ .

Intermediate

$x=-\tfrac{b}{2a}=2$ , then $y=4-8+1=-3$ :

(2,-3).

Find the discriminant of $x^{2}+2x+5$ and state the nature of the roots.

Intermediate

$\Delta=4-20=-16<0$ :

\text{no real roots (two complex)}.

Simplify $(3+2i)+(1-5i)$ .

Intermediate

4-3i.

Multiply $(2+i)(3-i)$ .

Intermediate

$6+i-i^{2}=6+i+1$ :

7+i.

Solve $x^{2}+4=0$ .

Advanced

$x^{2}=-4$ :

x=\pm 2i.

Solve $2x^{2}-4x-3=0$ with the quadratic formula.

Advanced

x=\frac{4\pm\sqrt{16+24}}{4}=\frac{2\pm\sqrt{10}}{2}.

State the degree of $4x^{3}-2x^{5}+7$ .

Basic

5.

Describe the end behaviour of $y=-x^{3}$ as $x\to+\infty$ .

Intermediate

y\to-\infty.

Factor $x^{3}-9x$ completely.

Intermediate

x(x-3)(x+3).

Find the remainder when $x^{2}+3x-4$ is divided by $(x-1)$ .

Intermediate

Evaluate at $x=1$ : $1+3-4$ :

0.

Find the zeros of $(x-2)(x+1)(x-5)$ .

Advanced

x=2,\ -1,\ 5.

The cubic $x^{3}-6x^{2}+11x-6$ factors as $(x-1)(x-2)(x-3)$ . State its roots.

Advanced

x=1,\ 2,\ 3.

Simplify $\dfrac{x^{2}-9}{x-3}$ .

Basic

Factor and cancel ( $x\neq 3$ ):

x+3.

Simplify $\dfrac{x}{2}+\dfrac{x}{3}$ .

Intermediate

\frac{3x+2x}{6}=\frac{5x}{6}.

State the vertical asymptote of $y=\dfrac{1}{x-4}$ .

Intermediate

x=4.

State the horizontal asymptote of $y=\dfrac{2x}{x+1}$ .

Intermediate

Ratio of leading coefficients:

y=2.

Solve $\dfrac{1}{x}+\dfrac{1}{2}=1$ .

Advanced

$\dfrac{1}{x}=\dfrac12$ :

x=2.

Simplify $\dfrac{x+1}{x^{2}-1}$ .

Advanced

$x^{2}-1=(x-1)(x+1)$ , cancel $(x+1)$ :

\frac{1}{x-1}.

Evaluate $\log_{2}8$ .

Basic

$2^{3}=8$ :

3.

Evaluate $\log_{10}1000$ .

Basic

3.

Solve $2^{x}=16$ .

Intermediate

2^{x}=2^{4}\ \Rightarrow\ x=4.

Solve $\log_{3}x=2$ .

Intermediate

$x=3^{2}$ :

x=9.

Evaluate $\log 50+\log 2$ (base 10).

Intermediate

Product rule: $\log(50\times 2)=\log 100$ :

2.

Evaluate $\ln e^{5}$ .

Intermediate

5.

Solve $3\cdot 2^{x}=24$ .

Advanced

$2^{x}=8=2^{3}$ :

x=3.

Condense $2\log x+\log y$ .

Advanced

\log\!\big(x^{2}y\big).

Simplify $\sqrt{50}$ .

Basic

\sqrt{25\cdot 2}=5\sqrt{2}.

Evaluate $8^{2/3}$ .

Intermediate

\big(\sqrt[3]{8}\big)^{2}=4.

Simplify $\sqrt{12}+\sqrt{27}$ .

Intermediate

2\sqrt{3}+3\sqrt{3}=5\sqrt{3}.

Solve $\sqrt{x+3}=5$ .

Intermediate

Square: $x+3=25$ :

x=22.

Rationalise $\dfrac{1}{\sqrt{2}}$ .

Advanced

\frac{1}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{\sqrt2}{2}.

Evaluate $16^{3/4}$ .

Advanced

\big(\sqrt[4]{16}\big)^{3}=2^{3}=8.

Write the next term of $3,\,7,\,11,\,15,\ldots$

Basic

Common difference 4:

19.

Find the 10th term of the arithmetic sequence with $a_1=2$ , $d=3$ .

Intermediate

2+9(3)=29.

Find the common ratio of $2,\,6,\,18,\,54,\ldots$

Intermediate

r=3.

Find the sum $1+2+4+8+16$ .

Intermediate

31.

Find the sum of the infinite series $8+4+2+1+\cdots$

Advanced

Geometric, $r=\tfrac12$ :

S=\frac{8}{1-\tfrac12}=16.

Evaluate $\displaystyle\sum_{k=1}^{4}(2k+1)$ .

Advanced

3+5+7+9=24.

Give the centre and radius of $x^{2}+y^{2}=25$ .

Basic

\text{centre } (0,0),\quad r=5.

Give the centre and radius of $(x-2)^{2}+(y+3)^{2}=16$ .

Intermediate

\text{centre } (2,-3),\quad r=4.

Is $\dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1$ an ellipse or a hyperbola?

Intermediate

The two terms are added:

\text{ellipse}.

Find the vertices of $\dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1$ .

Advanced

Opens left/right with $a=4$ :

(\pm 4,\,0).

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