Precalculus Transformations of Graphs Worksheet Solutions (PDF)

Precalculus math worksheet solutions: Transformations of Graphs – Maths PDF

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This precalculus transformations answers for Precalculus covers transformations of graphs (Common Core F-BF.B.3, F-IF.C.7b), with a free PDF to print.

Use this key to check your work. Each solution names the transformation or shows the substitution, and final answers are in bold.

Precalculus · Common Core State Standards: F-BF.B.3, F-IF.C.7b

Precalculus transformations answers: exercises 1 to 10

Exercise 1 – Read a horizontal shift

$x+3=x-(-3)$, so $h=-3$.

Shifted left 3 units

Exercise 2 – Read a vertical shift

The $-5$ is added to the output.

Shifted down 5 units

Exercise 3 – Reflection

The negative sign is outside the function, so it changes the sign of $y$.

Reflected over the $x$-axis

Exercise 4 – Vertical stretch

Every output is multiplied by 2.

Vertical stretch by a factor of 2

Exercise 5 – Move a point

Add 4 to $x$ and 1 to $y$: $(2+4,\ 5+1)$.

$(6,6)$

Exercise 6 – Reflect and shift a point

$x\to 3-1=2$. $y\to -(-2)+3=5$.

$(2,5)$

Exercise 7 – Write an equation

Reflect: $y=-x^{2}$. Shift right 3: $y=-(x-3)^{2}$. Shift down 2.

$y=-(x-3)^{2}-2$

Exercise 8 – Vertex of a transformed parabola

In $a(x-h)^{2}+k$, $h=-1$ and $k=-6$. Since $a=2>0$, it opens upward.

Vertex $(-1,-6)$; opens up

Exercise 9 – Identify from a graph

The vertex moved from (0, 0) to (2, -3).

The vertex is at $(2,-3)$, so shift right 2 and down 3.

$y=|x-2|-3$

Exercise 10 – Horizontal changes of a point

For $f(2x)$, the $x$-values are halved: $(2,6)$. For $f(x/3)$, the $x$-values are tripled: $(9,-1)$.

$(2,6)$ and $(9,-1)$

Exercises 11 and up

Exercise 11 – Even or odd?

(a) $f(-x)=x^{4}-3x^{2}=f(x)$: even. (b) $g(-x)=-x^{3}-x=-g(x)$: odd. (c) $h(-x)=x^{2}-x$, which is neither $h(x)$ nor $-h(x)$.

(a) even; (b) odd; (c) neither

Exercise 12 – Sequence of transformations

Shift right 2, stretch vertically by 3, reflect over the $x$-axis, shift up 1. At $x=6$: $-3\sqrt{4}+1=-6+1=-5$.

Value $-5$

Exercise 13 – Domain and range after shifts

$x+4\ge 0$, so $x\ge -4$. The root is at least 0, so $y\ge -3$.

Domain $[-4,\infty)$; range $[-3,\infty)$

Exercise 14 – Transform a table

Values of g.

Add 2 to each $x$ and 1 to each $y$.

See the table

Exercise 15 – Reflect over the y-axis

Need $-x\ge 0$, so $x\le 0$. Reflect over the $y$-axis: $(-9,3)$.

Domain $(-\infty,0]$; point $(-9,3)$

Exercise 16 – Equation from a graph

Vertex and one point.

$h=2,\ k=3$. Use $(3,2)$: $2=a(1)^{2}+3$, so $a=-1$.

$y=-(x-2)^{2}+3$

Exercise 17 – Fountain

Top at (4, 8); landing at (8, 0).

Maximum: $k=8$ ft at $x=4$. Set $y=0$: $(x-4)^{2}=16$, so $x=0$ or $x=8$. The graph is $y=x^{2}$ reflected over the $x$-axis, compressed by $\frac12$, shifted right 4 and up 8.

$8$ ft; lands at $x=8$ ft

Exercise 18 – Taxi fare transformation

$N(m)=0.8f(m)+4$: vertical compression by $0.8$, then shift up 4. $N(10)=0.8(25)+4=24$.

Compress by $0.8$, shift up 4; $N(10)=\$24$

Exercise 19 – Who is right?

$|3x+2|=|3(x+\frac23)|$, so it shifts left $\frac23$, not 2. For a shift left 2 after compressing, use $y=|3(x+2)|=|3x+6|$.

No; the correct equation is $y=|3(x+2)|$

Exercise 20 – All features of a transformed parabola

Graph of g.

Vertex $(-1,5)$; since $a< 0$, the range is $(-\infty,5]$. $g(0)=-2+5=3$. Solve $-2(x+1)^{2}+5=0$: $(x+1)^{2}=2.5$, $x=-1\pm\sqrt{2.5}\approx 0.58,\ -2.58$.

Vertex $(-1,5)$; range $(-\infty,5]$; $y$-int $3$; $x$-int about $0.58$ and $-2.58$

Key facts behind this precalculus transformations answers

Key facts

f(x) + k shifts up k; f(x – h) shifts right h.
-f(x) reflects over the x-axis; f(-x) reflects over the y-axis.
a f(x) stretches vertically by |a|; f(bx) compresses horizontally by 1/|b|.
A point (p, q) on f becomes (p + h, a q + k) on a f(x – h) + k.
Even functions satisfy f(-x) = f(x); odd functions satisfy f(-x) = -f(x).

How to use this precalculus transformations answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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