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Transformations of Graphs: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Tracing a point back

The point $(-1,7)$ on $y=-3f(x+2)+4$ comes from a point on $y=f(x)$ . Find that original point.
1. Hint 1
  The original input is the argument fed to $f$ , and the original output must undo the outside operations.
2. Hint 2
  Substitute the point's input into the inside expression, then solve the outside equation for the original height.
Answer

$(1,-1)$ .

Full solution

The original input is $-1+2=1$ .
The original output satisfies
$\begin{aligned}-3v+4&=7\\-3v&=3\\v&=-1\end{aligned}$

Thus the point is $(1,-1)$ .
Moving it forward shifts its input to $-1$ and its height to
$-3(-1)+4=7.$

Answer

$(1,-1)$ .

Key idea

Tracing a transformed point backward undoes the outside operations and evaluates the inside argument.
Problem 1: Mark complete
Problem 2 A shifted zero

The only real zero of $f$ is $5$ . Find the zero of $g(x)=f(3x+1)$ .
1. Hint 1
  The new function is zero when its input to $f$ reaches the original zero.
2. Hint 2
  Set the complete inside argument equal to the input where $f$ vanishes.
Answer

$x=\frac43$ .

Full solution

The only way to obtain output zero is
$\begin{aligned}3x+1&=5\\3x&=4\\x&=\frac43\end{aligned}$

The inside is a one-to-one linear expression, so this yields exactly one zero of $g$ .

Answer

$x=\frac43$ .

Key idea

An inside transformation moves a known zero by solving for the input that reaches it.
Problem 2: Mark complete
Problem 3 Three allowed locations

The domain of $f$ is $\{-4,0,6\}$ . Find the domain of $g(x)=f(-2x+2)$ .
1. Hint 1
  The inside argument must equal one of the three allowed original inputs.
2. Hint 2
  Solve $-2x+2=u$ for each $u$ in the original domain.
Answer

$\{-2,1,3\}$ .

Full solution

The original input $u$ appears when
$x=1-\frac u2.$
For $u=-4$ , $x=3$ ; for $u=0$ , $x=1$ ; and for $u=6$ , $x=-2$ .
Thus the new domain is
$\{-2,1,3\}.$

No other inside value is allowed, so the new domain is exactly $\{-2,1,3\}$ .

Answer

$\{-2,1,3\}$ .

Key idea

Transforming a finite domain means applying the input correspondence to each allowed input.
Problem 3: Mark complete
Problem 4 A bent path

The figure shows the entire graph of $f$ . Draw $g(x)=-f(2x)+1$ on the same grid and give its domain and range.

The original graph, $f$ .

Text description of this figure

A grid with the horizontal axis labeled x running from -3 to 5 and the vertical axis labeled y running from -3 to 4, gridlines and number labels at every whole number, and the origin labeled 0. The graph labeled f is two straight segments: one rising from the filled point (-2, 1) to the filled point (0, 3), and one falling from (0, 3) to the filled point (4, 1). The rest of the grid is blank, for the student to draw a second graph.
1. Hint 1
  Apply the same input and output changes to every point of the two segments.
2. Hint 2
  Divide each old horizontal coordinate by $2$ , and replace each old height $v$ by $1-v$ .
Answer

Segments through $(-1,0)$ , $(0,-2)$ , $(2,0)$ ; domain $[-1,2]$ , range $[-2,0]$ .

Full solution

The three corners of the original graph are $(-2,1)$ , $(0,3)$ , and $(4,1)$ .
Their images are
$\begin{aligned}(-2,1)&\mapsto(-1,0)\\(0,3)&\mapsto(0,-2)\\(4,1)&\mapsto(2,0)\end{aligned}$

Join the image points by straight segments and include the endpoints.
The horizontal coverage is $[-1,2]$ and the vertical coverage is $[-2,0]$ .

Answer

Segments through $(-1,0)$ , $(0,-2)$ , $(2,0)$ ; domain $[-1,2]$ , range $[-2,0]$ .

Key idea

A transformation of straight segments is determined by transforming their endpoints and corners.
Problem 4: Mark complete
Problem 5 Matching two procedures

One procedure shifts $y=f(x)$ upward by 4 and then multiplies all heights by 3. A second multiplies heights by 3 and then shifts upward by $k$ . Find $k$ so the procedures give the same graph for every function $f$ with nonempty domain.
1. Hint 1
  The shift in the first procedure is also affected by the later scaling.
2. Hint 2
  Write the resulting height from an arbitrary original height $v$ in each procedure.
Answer

$k=12$ .

Full solution

The first procedure sends $v$ to $3(v+4)$ , while the second sends it to $3v+k$ .
Equality requires
$\begin{aligned}3v+12&=3v+k\\k&=12\end{aligned}$

With this value the two height rules agree at every point, and neither procedure changes inputs.

Answer

$k=12$ .

Key idea

Reordering a shift and a scale requires adjusting the shift to compensate.
Problem 5: Mark complete
Problem 6 A flattened curve

A function $f$ has domain $(-2,4]$ and range $[1,7]$ . The rule $g(x)=0\cdot f(2x-2)+3$ retains the requirement that its argument be in the domain of $f$ . Find the domain and range of $g$ .
1. Hint 1
  Zeroing the output does not make an undefined inner evaluation valid.
2. Hint 2
  Solve the original domain condition on $2x-2$ , then apply the outside rule to every allowed output.
Answer

Domain $(0,3]$ ; range $\{3\}$ .

Full solution

The inside condition is
$-2<2x-2\le4.$
Adding $2$ and dividing by $2$ gives
$0<x\le3.$

At every allowed input, the finite output of $f$ is multiplied by zero and then increased by $3$ , so every output is $3$ .
The excluded inputs remain excluded.

Answer

Domain $(0,3]$ ; range $\{3\}$ .

Key idea

A zero outside multiplier collapses the range while preserving the inside domain restriction.
Problem 6: Mark complete
Problem 7 An input with a scale

Let $f$ have domain $[-4,8)$ and range $(-2,6]$ . Describe the horizontal moves for $g(x)=-2f(4x+8)+1$ , then give its domain and range.
1. Hint 1
  Factor the inside before reading its shift, and handle the output separately.
2. Hint 2
  Write $4x+8=4(x+2)$ ; a negative outside factor reverses the order of range endpoints.
Answer

Divide horizontal coordinates by $4$ , then shift left $2$ ; domain $[-3,0)$ ; range $[-11,5)$ .

Full solution

The inside factors as $4(x+2)$ , so old input $u$ moves to $u/4-2$ .
The endpoints $-4$ and $8$ move to $-3$ and $0$ with their inclusion unchanged, giving $[-3,0)$ .

An old output $v$ becomes $1-2v$ .
At $v=6$ this is $-11$ , included; as $v$ approaches the excluded lower value $-2$ , the output approaches $5$ , excluded.
Thus the range is $[-11,5)$ .

Answer

Divide horizontal coordinates by $4$ , then shift left $2$ ; domain $[-3,0)$ ; range $[-11,5)$ .

Key idea

Factoring the inside reveals the shift, and negative output scaling exchanges the upper and lower range bounds.
Problem 7: Mark complete
Problem 8 Right, then down

A student says shifting a graph right by 2 and shifting it down by 5 give the same final graph in either order. Is this true for any real-valued function? Explain using a general graph point.
1. Hint 1
  The moves affect separate coordinates.
2. Hint 2
  Track an arbitrary point $(u,v)$ through each order.
Answer

Yes; both orders send $(u,v)$ to $(u+2,v-5)$ .

Full solution

Right then down sends $(u,v)$ first to $(u+2,v)$ and then to $(u+2,v-5)$ .
Down then right sends it first to $(u,v-5)$ and then to the same final point.

Both rules are
$g(x)=f(x-2)-5.$
They have the same shifted domain and the same output at every allowed input.

Answer

Yes; both orders send $(u,v)$ to $(u+2,v-5)$ .

Key idea

Moves affecting separate coordinates can commute even though two moves on one coordinate may not.
Problem 8: Mark complete
Problem 9 An odd rule moved

Let $f$ be odd on all real numbers, and define $g(x)=-f(-x)+5$ . A student claims $g$ is also odd. Decide whether that can be true and justify your answer.
1. Hint 1
  Use the defining identity of an odd function before analyzing the added constant.
2. Hint 2
  An odd function defined at zero must have output zero there.
Answer

False; $g(0)=5$ .

Full solution

Oddness gives $f(-x)=-f(x)$ , so
$g(x)=f(x)+5.$
Also,
$\begin{aligned}f(0)&=-f(0)\\f(0)&=0\end{aligned}$

Thus $g(0)=5$ .
An odd function on all real numbers must have value zero at zero, so $g$ cannot be odd.

Answer

False; $g(0)=5$ .

Key idea

A transformation that preserves one symmetry step can still lose that symmetry after a nonzero vertical shift.
Problem 9: Mark complete
Problem 10 Matching formulas

Let $f(x)=\lvert x\rvert$ on the declared domain $[-1,1]$ . A student says $g(x)=f(2x)$ and $h(x)=2f(x)$ are equal functions because both simplify to $2\lvert x\rvert$ . Take both codomains to be $[0,2]$ . Assess the claim.
1. Hint 1
  Equal formulas need equal domains as well as equal codomains.
2. Hint 2
  Compare the input requirements for evaluating $f(2x)$ and $f(x)$ .
Answer

False; $g$ has domain $[-\frac12,\frac12]$ , while $h$ has domain $[-1,1]$ .

Full solution

To evaluate $g$ , the inside must satisfy $-1\le2x\le1$ , giving
$-\frac12\le x\le\frac12.$

The domain of $h$ is still $[-1,1]$ .
The formulas agree wherever both are defined, but $h(1)=2$ while $g(1)$ is undefined.
Different domains make the functions unequal.

Answer

False; $g$ has domain $[-\frac12,\frac12]$ , while $h$ has domain $[-1,1]$ .

Key idea

Coinciding transformed formulas do not establish function equality unless the transformed domains also coincide.
Problem 10: Mark complete

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