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Stretching and Reflecting 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 Three recorded points

The graph shows the complete function $f$ . Draw $y=2f(x)$ on the same grid.

The complete graph of $f$ : three separate points.

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative three to four and the vertical y-axis from negative five to seven, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Three separate filled dots are plotted and no line joins them: one at negative two, negative two; one at one, zero, sitting on the x-axis; and one at three, three. Those three dots are the whole graph of f. No coordinates are printed beside the dots, and nothing else is drawn on the grid.
1. Hint 1
  An output factor changes heights and leaves each input fixed.
2. Hint 2
  Read and double every plotted height, preserving the finite domain.
Answer

Isolated points $(-2,-4)$ , $(1,0)$ , and $(3,6)$ .

Full solution

The original points are $(-2,-2)$ , $(1,0)$ , and $(3,3)$ .
Multiplying their heights by two gives $(-2,-4)$ , $(1,0)$ , and $(3,6)$ .
Plot these three points without connecting them.

The zero-height point remains fixed, while each nonzero height is twice as far from the horizontal axis.
This checks the requested vertical scaling.

Answer

Isolated points $(-2,-4)$ , $(1,0)$ , and $(3,6)$ .

Key idea

A vertical scaling multiplies every height by the same factor, so zero heights stay put and the input set is unchanged.
Problem 1: Mark complete
Problem 2 The relocated point

The point $(12,5)$ lies on the graph of $f$ . For a positive constant $b$ , this point reappears on the graph of $y=f(bx)$ at $x=3$ . Find $b$ , and check your value in the equation you solve.
1. Hint 1
  A factor inside the function leaves every height alone and moves only the input positions.
2. Hint 2
  At the new input, the inside expression has to equal the old input, which gives an equation for the factor.
Answer

$b=4$ ; check: $3(4)=12$ .

Full solution

The height $5$ returns where the inside expression equals the old input $12$ .
At the new input $3$ that requires
$3b=12,$
so $b=4$ .

Check: substituting back gives $3(4)=12$ , matching the old input, so $b=4$ is confirmed.

Answer

$b=4$ ; check: $3(4)=12$ .

Key idea

One point and its new position fix the inside factor, which divides every input while leaving every height alone.
Problem 2: Mark complete
Problem 3 The flipped heights

The points $(-5,2)$ , $(0,-3)$ , and $(3,7)$ lie on the graph of $f$ . Give the three corresponding points on the graph of $y=-f(x)$ .
1. Hint 1
  A minus sign outside the function acts on heights alone, so every input keeps its place.
2. Hint 2
  Change the sign of each second coordinate; the negative height $-3$ becomes $3$ .
Answer

$(-5,-2)$ , $(0,3)$ , and $(3,-7)$ .

Full solution

Negating an output keeps its input and reverses the sign of its height.
The three points therefore become $(-5,-2)$ , $(0,3)$ , and $(3,-7)$ , each the same distance from the horizontal axis as before but on the opposite side.

Check the first one: $f(-5)=2$ , so $-f(-5)=-2$ , which matches the point $(-5,-2)$ above.

Answer

$(-5,-2)$ , $(0,3)$ , and $(3,-7)$ .

Key idea

Reflecting across the horizontal axis negates every height while leaving every input exactly where it was.
Problem 3: Mark complete
Problem 4 The broken line

The graph shows all of $f$ . Draw $g(x)=-f(-x)$ , and give its domain and range.

The complete graph of $f$ .

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative four to four and the vertical y-axis from negative five to five, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Two solid straight segments join three points into one broken line: it rises from negative three, one to negative one, four, then falls to two, two. A filled dot marks each end and the corner, and the letter f labels the broken line. No coordinates are printed beside the points, and no other marks are on the grid.
1. Hint 1
  The inside negative changes horizontal positions, while the outside negative changes heights.
2. Hint 2
  Negate both coordinates of every endpoint and corner, then preserve the original connections.
Answer

Two segments, one from $(-2,-2)$ to $(1,-4)$ and one from $(1,-4)$ to $(3,-1)$ . Domain $-2\le x\le3$ ; range $-4\le y\le-1$ .

Full solution

The original points $(-3,1)$ , $(-1,4)$ , and $(2,2)$ map to $(3,-1)$ , $(1,-4)$ , and $(-2,-2)$ .
Connect corresponding points with solid segments.
Reading from left to right, the transformed graph begins at $(-2,-2)$ , dips to $(1,-4)$ , then rises to $(3,-1)$ .

The original domain $-3\le x\le2$ becomes $-2\le x\le3$ , and the original range $1\le y\le4$ becomes
$-4\le y\le-1.$
Negating the coordinates a second time would restore every original point.

Answer

Two segments, one from $(-2,-2)$ to $(1,-4)$ and one from $(1,-4)$ to $(3,-1)$ . Domain $-2\le x\le3$ ; range $-4\le y\le-1$ .

Key idea

Inside and outside negatives reflect a graph across different axes and together negate both coordinates.
Problem 4: Mark complete
Problem 5 An altered bowl

Let $f(x)=(x+2)^2-1$ on all real inputs. Write $g(x)=-2f(x-1)+3$ in vertex form, identify its vertex and opening direction, and draw its graph on the blank grid.

A blank grid on which to draw $g$ .

Text description of this figure

An empty coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative four to three and the vertical y-axis from negative five to seven, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Nothing is drawn on the grid beyond the two axes, their arrowheads, the tick marks, the whole-number labels and the gridlines.
1. Hint 1
  Substitute the new input into the whole existing rule before applying the outside operations.
2. Hint 2
  Keep the resulting square intact so the vertex and scale are visible.
Answer

$g(x)=-2(x+1)^2+5$ ; vertex $(-1,5)$ ; opens downward.

Full solution

Substituting into the old rule gives
$f(x-1)=(x+1)^2-1.$
Therefore
$g(x)=-2((x+1)^2-1)+3.$
Distributing outside the square produces
$g(x)=-2(x+1)^2+5.$

The vertex is $(-1,5)$ and the negative coefficient makes it a maximum.
Points one unit to either side have height $3$ , and points two units to either side have height $-3$ .
Plot $(-2,3)$ , $(0,3)$ , $(-3,-3)$ , and $(1,-3)$ with the vertex and draw the downward parabola through them.
Substitution confirms those heights.

Answer

$g(x)=-2(x+1)^2+5$ ; vertex $(-1,5)$ ; opens downward.

Key idea

Substituting the new input into the whole old rule is a reliable way to reach the new vertex form.
Problem 5: Mark complete
Problem 6 Two finite transformations

The complete records of $f$ are $f(0)=1$ , $f(2)=3$ , and $f(4)=2$ . Let $g(x)=f(2x)$ , defined for $x=0,1,2$ , and $h(x)=2f(x)$ , defined for $x=0,2,4$ . Find all points shared by the graphs of $g$ and $h$ .
1. Hint 1
  Compare both the input sets and the heights of the transformed finite graphs.
2. Hint 2
  List each transformed pair, then keep only pairs that occur in both complete records.
Answer

No points are shared.

Full solution

Horizontal compression gives the complete points of $g$ : $(0,1)$ , $(1,3)$ , and $(2,2)$ .
Vertical scaling gives the complete points of $h$ : $(0,2)$ , $(2,6)$ , and $(4,4)$ .

At common input zero the heights are $1$ and $2$ , and at common input two they are $2$ and $6$ .
No heights agree at any common input.
Therefore the graphs have no shared points.

Answer

No points are shared.

Key idea

Graphs share a point only when both its input and output agree.
Problem 6: Mark complete
Problem 7 The target height

The graph shows the complete function $f$ . Find every solution of $3f(2x)=6$ , and state the domain of $y=3f(2x)$ .

The complete graph of $f$ .

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative five to seven and the vertical y-axis from negative one to five, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Two solid straight segments form one broken line: it rises steadily from negative four, zero on the x-axis to zero, four on the y-axis, then falls more gently to six, one. A filled dot marks each of those three points, and this broken line is the whole graph of f. No coordinates are printed beside the points, and no other marks are on the grid.
1. Hint 1
  The outside factor determines which original height is needed, and the inside factor determines the new input positions.
2. Hint 2
  Read every original input at the required height, then divide those inputs by the inside factor.
Answer

Solutions: $x=-1$ and $x=2$ . Domain: $-2\le x\le3$ .

Full solution

Dividing the equation by the nonzero number three gives
$f(2x)=2.$
The original graph reaches height two at inputs $-2$ and $4$ .
Thus $2x=-2$ or $2x=4$ , giving $x=-1$ or $x=2$ .

Write $u=2x$ for the input handed to $f$ .
The graph gives the domain $-4\le u\le6$ , and requiring $u$ to stay in that interval gives
$-2\le x\le3.$
Each solution found above lies in that interval and produces original height two, which the outside factor changes to six.

Answer

Solutions: $x=-1$ and $x=2$ . Domain: $-2\le x\le3$ .

Key idea

A combined scaling changes the target height and the locations that produce it in separate steps.
Problem 7: Mark complete
Problem 8 The compared heights

A function satisfies $f(-1)=-3$ and $f(2)=0$ . Rae says every point of $y=4f(x)$ lies higher than its corresponding point on $y=f(x)$ . Is Rae correct? Use both given records to explain.
1. Hint 1
  Multiplication by a positive factor larger than one affects positive, negative, and zero heights differently.
2. Hint 2
  Compare the original and scaled heights at each supplied input.
Answer

No; the height at $x=-1$ changes from $-3$ to $-12$ , and the height at $x=2$ stays $0$ .

Full solution

At input $-1$ , the scaled height is
$4f(-1)=4(-3),$
which is $-12$ , lower than the original height $-3$ .
At input $2$ , the scaled height is
$4f(2)=4(0),$
which remains zero.

The first point moves down, and the second does not move at all.
These records disprove the claim.
A vertical stretch increases distance from the horizontal axis for nonzero heights, which is different from increasing every height.

Answer

No; the height at $x=-1$ changes from $-3$ to $-12$ , and the height at $x=2$ stays $0$ .

Key idea

A vertical stretch moves nonzero heights farther from the horizontal axis and leaves zero heights fixed.
Problem 8: Mark complete
Problem 9 A mirrored increasing graph

The function $f$ is strictly increasing on $1\le x\le4$ . Let $g(x)=f(-x)$ , defined for $-4\le x\le-1$ . Decide whether $g$ is strictly increasing or strictly decreasing on that domain, and justify your answer using two inputs from it.
1. Hint 1
  Negating two ordered inputs reverses their order before the original function acts.
2. Hint 2
  Choose $u<v$ in the new domain and compare $-u$ with $-v$ .
Answer

$g$ is strictly decreasing on $-4\le x\le-1$ .

Full solution

If $-4\le u<v\le-1$ , then both $-u$ and $-v$ are in the original domain, and $-u>-v$ .
Since $f$ is strictly increasing,
$f(-u)>f(-v).$
These values are $g(u)$ and $g(v)$ , so the larger new input $v$ gives the smaller output.

This holds for every such pair, proving strict decrease across the whole reflected domain.

Answer

$g$ is strictly decreasing on $-4\le x\le-1$ .

Key idea

A reflection across the vertical axis reverses input order and changes a strictly increasing graph into a strictly decreasing one.
Problem 9: Mark complete
Problem 10 The axis crossings

For a real constant $a$ , compare the $x$ -intercepts of $y=f(x)$ and $y=af(x)$ , using the same nonempty domain. Prove that the sets of $x$ -intercepts agree for every function when $a\ne0$ , and give a counterexample showing why the condition on $a$ matters.
1. Hint 1
  Start from what an $x$ -intercept is: an input whose output is zero, read for each of the two graphs.
2. Hint 2
  Argue both ways: every old zero output must survive the scaling, and the scaling must create no new one.
3. Hint 3
  For the second part, ask what every output of $af(x)$ becomes when $a$ is zero.
Answer

For $a\ne0$ , the intercept sets agree. For $a=0$ and $f(x)=1$ on all real inputs, $f$ has none but $af$ has every point $(x,0)$ .

Full solution

If $f(x)=0$ , then $af(x)=a(0)=0$ , so every old intercept remains.
Conversely, if $af(x)=0$ and $a\ne0$ , divide by that nonzero constant to obtain
$f(x)=0.$
Thus no new intercept is added.

The condition is necessary for a guarantee applying to every function.
With $a=0$ and $f(x)=1$ on all real inputs, the original graph has no horizontal-axis intercept, while the new output is $0(1)=0$ at every input.
The intercept sets then differ.

Answer

For $a\ne0$ , the intercept sets agree. For $a=0$ and $f(x)=1$ on all real inputs, $f$ has none but $af$ has every point $(x,0)$ .

Key idea

A nonzero vertical scale preserves exactly the zero outputs, while a zero scale collapses every output to zero.
Problem 10: Mark complete

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