Core practice ← Back to lesson

Shifting 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)

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 A total of two heights

A function satisfies $f(1)+f(4)=9$ . Let $g(x)=f(x)+3$ on the same domain. Find $g(1)+g(4)$ .
1. Hint 1
  The vertical change applies separately to each output in the total.
2. Hint 2
  Replace each value of the new function, then collect the two added constants.
Answer

$15$ .

Full solution

The two requested outputs are $f(1)+3$ and $f(4)+3$ .
Their sum is
$g(1)+g(4)=f(1)+f(4)+6.$
The given total is $9$ , so the new total is $9+6=15$ .
Each of the two graph heights rose by three.

Answer

$15$ .

Key idea

A vertical shift changes every selected height, so a total changes once for each output included.
Problem 1: Mark complete
Problem 2 Two successive rules

Let $g(x)=f(x-2)$ and $h(x)=g(x+5)$ . Write $h(x)$ directly in terms of $f$ , with a single simplified input expression.
1. Hint 1
  The input supplied to the second rule must replace the whole input of the first.
2. Hint 2
  Substitute the expression $x+5$ into the rule for $g$ , then combine the constants inside $f$ .
Answer

$h(x)=f(x+3)$ .

Full solution

The rule for $g$ subtracts $2$ from whatever input it receives.
Therefore
$h(x)=f((x+5)-2).$
Simplifying gives
$h(x)=f(x+3).$
The first translation is right two and the second left five, giving a net move left three.

Answer

$h(x)=f(x+3)$ .

Key idea

Successive horizontal translations combine through substitution of the entire new input.
Problem 2: Mark complete
Problem 3 The two shifted spans

The domain of $f$ is $-5\le x\le-1$ and its range is $0\le y\le6$ . A function $g$ is given by $g(x)=f(x-h)+k$ ; its domain is $-1\le x\le3$ and its range is $-4\le y\le2$ . Find $h$ and $k$ .
1. Hint 1
  The change inside the function moves the span of inputs, while the change outside moves the span of outputs.
2. Hint 2
  Match the left ends of the two domains for one equation, then match the lowest outputs for the other.
Answer

$h=4$ and $k=-4$ .

Full solution

The subtraction inside moves every input by $h$ , so the ends of the domain move to $-5+h$ and $-1+h$ .
Matching the left ends gives
$-5+h=-1.$
Hence $h=4$ , and the right ends agree, since $-1+4=3$ .

The addition outside moves every output by $k$ , so the ends of the range move to $0+k$ and $6+k$ .
Matching the lowest outputs gives
$0+k=-4.$
Hence $k=-4$ , and the highest outputs agree, since $6-4=2$ .
The graph moved right four and down four.

Answer

$h=4$ and $k=-4$ .

Key idea

A horizontal shift can be read from the domain and a vertical shift from the range, because each change touches only one of them.
Problem 3: Mark complete
Problem 4 The separated pieces

The graph shows the entire function $f$ . Draw $g(x)=f(x-1)-1$ on the same grid, then give every intercept of $g$ .

The complete graph of $f$ , in two separated pieces.

Text description of this figure

A coordinate grid with the horizontal axis x running from negative 4 to 6 and the vertical axis y running from negative 2 to 7, ruled with gridlines and numbered at every whole number, with the origin labeled 0. The graph of f, labeled f beside its upper piece, is two separate straight segments: one rises from the point (negative 1, 0) to the point (1, 2), and the other rises from the point (3, 4) to the point (5, 6). All four endpoints are filled dots, and nothing at all is drawn between the two segments. No second graph is drawn.
1. Hint 1
  Translate all endpoints and retain the gap between the pieces.
2. Hint 2
  A minus inside shifts inputs right; the outside subtraction lowers heights.
3. Hint 3
  Read intercepts from the completed shifted segments, not from the old intercept names.
Answer

$x$ -intercept $(1,0)$ ; $y$ -intercept $(0,-1)$ .

Full solution

Each original point moves one unit right and one unit down.
The segments therefore run from $(0,-1)$ to $(2,1)$ and from $(4,3)$ to $(6,5)$ .
Draw those segments with included endpoints and leave the gap empty.

The first segment meets height zero at $x=1$ , so $(1,0)$ is the only horizontal-axis intercept; the second segment stays between heights $3$ and $5$ and never reaches zero.
That same first segment also includes input $x=0$ , where its height is $-1$ , so $(0,-1)$ is the vertical-axis intercept.
The change of each endpoint is exactly the same translation.

Answer

$x$ -intercept $(1,0)$ ; $y$ -intercept $(0,-1)$ .

Key idea

A translation preserves gaps while moving the domain, range, and possible axis meetings.
Problem 4: Mark complete
Problem 5 A delayed trace

The graph shows a temperature trace $f(t)$ for $0\le t\le4$ , where $t$ is measured in seconds and outputs are in degrees Celsius. A new recording starts this same trace $2$ seconds later and adds $1$ degree Celsius to each recorded temperature. Write its rule $g(t)$ in terms of $f$ , draw it on the grid, and state its domain and range.

The original recorded trace $f(t)$ .

Text description of this figure

A coordinate grid whose horizontal axis is time in seconds, numbered from 0 to 7, and whose vertical axis is temperature in degrees Celsius, numbered from 0 to 6, with gridlines at every whole number. The trace labeled f is a broken line of three straight pieces: it rises from the point (0, 2) to the point (1, 4), falls to the point (3, 1), then rises to the point (4, 3). Each of those four points is a filled dot, and no other trace is drawn on the grid.
1. Hint 1
  The delay moves the same trace to later input times, while the temperature adjustment changes outputs.
2. Hint 2
  To reproduce an old reading at time $u$ , determine the new time and new temperature of that point.
Answer

$g(t)=f(t-2)+1$ ; domain $2\le t\le6$ seconds; range $2\le g(t)\le5$ degrees Celsius.

Full solution

The reading formerly at time $u$ appears at time $u+2$ , so the input passed to $f$ must be $t-2$ .
The temperature adjustment is outside the function, giving
$g(t)=f(t-2)+1.$

Translate the original corners to $(2,3)$ , $(3,5)$ , $(5,2)$ , and $(6,4)$ and join them in order.
The time span becomes $2\le t\le6$ seconds, and the original temperature span from $1$ to $4$ becomes $2$ to $5$ degrees Celsius.
At new time $2$ , the output is the old starting value $2$ plus $1$ , which checks the start.

Answer

$g(t)=f(t-2)+1$ ; domain $2\le t\le6$ seconds; range $2\le g(t)\le5$ degrees Celsius.

Key idea

A delayed record needs a subtraction inside its function so the old input occurs at a later time.
Problem 5: Mark complete
Problem 6 A graph clear of the axis

The complete graph of $f$ is shown. Find every real $k$ for which $y=f(x)+k$ has no $x$ -intercepts.

The complete graph of $f$ .

Text description of this figure

A coordinate grid with the horizontal axis x running from negative 4 to 5 and the vertical axis y running from negative 3 to 4, ruled with gridlines and numbered at every whole number, with the origin labeled 0. The graph shown is a broken line of three straight segments: it rises from the point (negative 3, negative 2) to the point (negative 1, 3), falls to the point (2, negative 1), then rises to the point (4, 2). Each of those four points is a filled dot. Nothing else is drawn: the highest and lowest points of the graph carry no labels, and no shifted copy of the graph appears.
1. Hint 1
  A vertical shift has an intercept precisely when one of its shifted heights equals zero.
2. Hint 2
  Read the full original range, then decide which vertical shifts place that entire range above or below zero.
Answer

$k>2$ or $k<-3$ .

Full solution

The original graph reaches every height from $-2$ to $3$ .
After the shift, the range extends from $k-2$ to $k+3$ .
It avoids zero exactly when the lower endpoint is positive or the upper endpoint is negative.

These conditions are
$k-2>0$
or
$k+3<0.$
They give $k>2$ or $k<-3$ .
At either boundary $k=2$ or $k=-3$ , an extreme point sits on the axis, so equality is excluded.

Answer

$k>2$ or $k<-3$ .

Key idea

A vertical shift avoids horizontal-axis intercepts when its entire range lies strictly on one side of zero.
Problem 6: Mark complete
Problem 7 The point at the origin

A graph is moved right $a$ units and up $b$ units, then right $b$ units and down $a$ units, where $a$ and $b$ are positive. The final position of one original point $(0,0)$ is $(5,-1)$ . Find $a$ and $b$ , and write the final rule in terms of the original function $f$ .
1. Hint 1
  The two horizontal displacements add, while the vertical displacements have opposite signs.
2. Hint 2
  Use the final coordinates to form two equations for the two positive distances.
Answer

$a=3$ , $b=2$ ; final rule $y=f(x-5)-1$ .

Full solution

The final horizontal position gives $a+b=5$ , while the vertical position gives $b-a=-1$ .
Adding these equations gives
$2b=4,$
hence $b=2$ and $a=3$ .
Both are positive, as required.

The total move is right five and down one, so the final graph has rule
$y=f(x-5)-1.$
Checking the original point sends $(0,0)$ to $(3,2)$ and then to $(5,-1)$ , the stated final position.

Answer

$a=3$ , $b=2$ ; final rule $y=f(x-5)-1$ .

Key idea

Tracking the two coordinates separately turns successive translations into equations for their displacements.
Problem 7: Mark complete
Problem 8 The order of two moves

A graph is shifted left $4$ units and up $2$ units. Lee says the final graph is unchanged if those two moves are performed in the opposite order. Is Lee correct for every function? Justify your answer by following an arbitrary point.
1. Hint 1
  One move changes the input coordinate and the other changes the output coordinate.
2. Hint 2
  Track a point through both orders and compare its final coordinates.
Answer

Yes; both orders send $(p,q)$ to $(p-4,q+2)$ and give $y=f(x+4)+2$ .

Full solution

Start with any graph point $(p,q)$ .
Moving left first sends it to $(p-4,q)$ , then moving up sends it to $(p-4,q+2)$ .
Moving up first gives $(p,q+2)$ , and moving left next also gives $(p-4,q+2)$ .

The arbitrary point has the same final position in both orders, so every graph point does.
Both descriptions give
$y=f(x+4)+2.$
Lee is correct for every function.

Answer

Yes; both orders send $(p,q)$ to $(p-4,q+2)$ and give $y=f(x+4)+2$ .

Key idea

Horizontal and vertical translations can be reordered because they change separate coordinates.
Problem 8: Mark complete
Problem 9 A visibly different graph

Mara claims shifting a function’s graph right by a positive distance must produce a different set of plotted points. Is this true for every function on all real inputs? Give a counterexample if it is false.
1. Hint 1
  Distinguish moving each individual point from changing the entire collection of points.
2. Hint 2
  Ask whether a moved point could land on a place the graph already occupied.
Answer

False; for example, $f(x)=5$ on all real inputs has $f(x-h)=5$ for every $h>0$ .

Full solution

Take the constant function $f(x)=5$ on all real inputs.
For any positive $h$ , substituting a different input changes no output:
$f(x-h)=5.$
Both graphs are the entire horizontal line at height $5$ .

Each particular point translates to a different point on that same line, but the complete set of points stays unchanged.
This disproves the claim.

Answer

False; for example, $f(x)=5$ on all real inputs has $f(x-h)=5$ for every $h>0$ .

Key idea

A translation can move individual points while leaving the complete graph of a constant function unchanged.
Problem 9: Mark complete
Problem 10 The peak and the crossing

The graph of $y=f(x)$ crosses the $x$ -axis at $x=-5$ , and its only highest point sits at input $x=-2$ . For a positive number $h$ , the graph of $y=f(x-h)$ has its only highest point at input $x=4$ . Find $h$ , give the new position of that crossing, and compare the distance between those two features before and after the change.
1. Hint 1
  Each marked feature travels with the graph, so this change moves its input while its height stays put.
2. Hint 2
  Measure the move from the highest point, then apply that same move to the crossing.
Answer

$h=6$ ; that crossing moves to $(1,0)$ ; the two features are $3$ apart both before and after.

Full solution

A change inside the function moves every point horizontally by the same amount, so the highest point keeps its height while its input grows from $-2$ to $-2+h$ .
That new input is $4$ , so
$-2+h=4,$
which gives $h=6$ .

The crossing is a point of the graph too, so it moves by that same $6$ , from $x=-5$ to
$-5+6=1.$
Its height is still $0$ , so it now sits at $(1,0)$ .

Before the change the two features are $-2-(-5)=3$ apart, and afterwards they are $4-1=3$ apart.
Both inputs grew by the same amount, so their difference cannot change: the shift carries the whole graph rigidly, spacing included.

Answer

$h=6$ ; that crossing moves to $(1,0)$ ; the two features are $3$ apart both before and after.

Key idea

A shift adds the same amount to every input, so marked features travel together and the distances between them are unchanged.
Problem 10: Mark complete

← Back to lesson