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Graphs of Functions: 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 The marked record

The marked point $P$ on the graph records the equation $f(r-1)=3$ . Find $r$ from this point.

Points recorded on the graph of $f$ , with $P$ marked.

Text description of this figure

A coordinate grid with the horizontal axis labeled x running from negative 4 to 4 and the vertical axis labeled y running from negative 2 to 5. Gridlines and number labels sit at every whole number, and the origin is labeled 0. Three filled dots are plotted and none of them are joined: one at (negative 2, 3), one at (0, negative 1) and one at (3, 2). Only the dot at (negative 2, 3) carries a label, the letter P; no coordinates are printed beside any dot.
1. Hint 1
  Every point of a graph pairs an input with the height above it, so $P$ supplies both numbers in $f(\text{input})=\text{output}$ .
2. Hint 2
  The first coordinate of the marked point is the entire input inside the function notation.
3. Hint 3
  Set the expression for that input equal to the point’s horizontal coordinate.
Answer

$r=-1$ .

Full solution

The marked point is $(-2,3)$ , so the input producing its height is $-2$ .
Therefore
$r-1=-2.$
Adding $1$ gives $r=-1$ .
Substituting back gives input $-1-1=-2$ , the horizontal position of $P$ .

Answer

$r=-1$ .

Key idea

An expression inside function notation must match the graph point’s input coordinate.
Problem 1: Mark complete
Problem 2 A change in height

The figure shows the complete graph of $g$ . Find $g(3)-g(-1)$ .

The complete graph of $g$ .

Text description of this figure

A coordinate grid with the horizontal axis labeled x running from negative 3 to 5 and the vertical axis labeled y running from negative 2 to 5. Gridlines and number labels sit at every whole number, and the origin is labeled 0. The whole graph of g is three joined straight segments: it rises from the endpoint (negative 2, 2) to the corner (0, 4), falls steeply to the corner (2, negative 2), then rises to the endpoint (4, 0). Filled dots mark the two endpoints and the two corners. No coordinates are printed and no guide lines are drawn.
1. Hint 1
  A change in output compares two heights at their specified inputs.
2. Hint 2
  Read each height separately, then subtract in the stated order.
Answer

$-4$ .

Full solution

At input $3$ , the height is $-1$ , while at input $-1$ the height is $3$ .
Thus
$g(3)-g(-1)=-1-3,$
which is $-4$ .
The negative result agrees with the lower final height.

Answer

$-4$ .

Key idea

Differences of function values compare output heights while preserving the order of the inputs.
Problem 2: Mark complete
Problem 3 The finite record

A table lists all inputs and outputs of $h$ : input $-3$ has output $2$ , input $-1$ has output $-2$ , input $2$ has output $0$ , and input $4$ has output $2$ . Plot the complete graph on the blank grid.

A blank grid on which to plot the graph of $h$ .

Text description of this figure

An empty coordinate grid with the horizontal axis labeled x running from negative 4 to 5 and the vertical axis labeled y running from negative 3 to 3. Gridlines and number labels sit at every whole number, and the origin is labeled 0. Nothing at all is plotted on the grid: no points, no segments and no curve.
1. Hint 1
  A complete table pairs each input with exactly its own output.
2. Hint 2
  The domain contains just the listed inputs, so decide whether any connecting points belong.
Answer

Four isolated points: $(-3,2)$ , $(-1,-2)$ , $(2,0)$ , $(4,2)$ .

Full solution

The complete table gives the pairs $(-3,2)$ , $(-1,-2)$ , $(2,0)$ , and $(4,2)$ .
Plot each with input horizontal and output vertical.

There are no other allowed inputs, so leave the points unconnected.
A segment would add infinitely many pairs absent from the table.
Each plotted point reproduces its corresponding input-output record.

Answer

Four isolated points: $(-3,2)$ , $(-1,-2)$ , $(2,0)$ , $(4,2)$ .

Key idea

A complete finite table produces isolated graph points unless additional inputs are explicitly allowed.
Problem 3: Mark complete
Problem 4 Reading one height

The graph shows all of $f$ . Find every solution of $f(x)=2$ , and state the $y$ -intercept.

The complete graph of $f$ .

Text description of this figure

A coordinate grid with the horizontal axis labeled x running from negative 4 to 5 and the vertical axis labeled y running from negative 2 to 4. Gridlines and number labels sit at every whole number, and the origin is labeled 0. The whole graph of f is three joined straight segments: it rises from the endpoint (negative 3, negative 2) to the corner (negative 1, 2), then runs exactly level to the corner (2, 2), then falls to the endpoint (4, negative 2). Filled dots mark the two endpoints and the two corners. No coordinates are printed, and no horizontal guide line is drawn.
1. Hint 1
  A horizontal piece can supply more solutions than isolated crossings do.
2. Hint 2
  Trace every location at height two, then separately find the point with input zero.
Answer

$f(x)=2$ for $-1\le x\le2$ ; $y$ -intercept $(0,2)$ .

Full solution

The first segment reaches height $2$ at $x=-1$ .
The entire middle segment stays at that height until $x=2$ .
The final segment then falls away from it.
Thus every input in
$-1\le x\le2$
solves the equation, and no other input does.

The input $0$ lies in that interval, so $f(0)=2$ .
The graph meets the vertical axis at $(0,2)$ .

Answer

$f(x)=2$ for $-1\le x\le2$ ; $y$ -intercept $(0,2)$ .

Key idea

An equation asking for a graph’s height can have a whole interval of solutions.
Problem 4: Mark complete
Problem 5 Two separate pieces

The figure shows the entire graph of $q$ . Give its domain, range, and every intercept.

The complete graph of $q$ , drawn in two pieces.

Text description of this figure

A coordinate grid with the horizontal axis labeled x running from negative 5 to 4 and the vertical axis labeled y running from negative 3 to 6. Gridlines and number labels sit at every whole number, and the origin is labeled 0. The whole graph of q is two separate straight segments with an empty gap between them. The left segment rises from the filled endpoint (negative 4, negative 2) to the filled endpoint (negative 1, 1). The right segment rises from the filled endpoint (1, 3) to the filled endpoint (3, 5). Nothing is drawn between the two pieces, no intercept is marked and no coordinates are printed.
1. Hint 1
  The gap matters when you collect the horizontal and vertical positions covered.
2. Hint 2
  Inspect each piece’s span and its meetings with the axes, without filling in the gap.
Answer

Domain: $-4\le x\le-1$ or $1\le x\le3$ . Range: $-2\le y\le1$ or $3\le y\le5$ . One $x$ -intercept, $(-2,0)$ , and no $y$ -intercept.

Full solution

The left piece covers inputs from $-4$ through $-1$ and heights from $-2$ through $1$ .
The right piece covers inputs from $1$ through $3$ and heights from $3$ through $5$ .
These separate spans give the stated domain and range.

The left segment rises one unit for every unit to the right, so it reaches height zero at $x=-2$ .
Thus $(-2,0)$ is its sole horizontal-axis intercept.
The right piece stays above the axis.
Input zero is absent, so there is no vertical-axis intercept.

Answer

Domain: $-4\le x\le-1$ or $1\le x\le3$ . Range: $-2\le y\le1$ or $3\le y\le5$ . One $x$ -intercept, $(-2,0)$ , and no $y$ -intercept.

Key idea

A graph with gaps can have separate domain and range intervals and can lack a vertical-axis intercept.
Problem 5: Mark complete
Problem 6 A missing height

A graph joins the listed points in order with straight segments: $(-3,1)$ , $(-1,k)$ , $(2,0)$ , and $(4,3)$ . Is there an integer $k$ that makes the function strictly decrease from $x=-3$ to $x=2$ and strictly increase from $x=2$ to $x=4$ ? If so, give $k$ , draw the completed graph, and identify its turning point; otherwise, explain why no such completion is possible.

The three given points, with the height at $x=-1$ left out.

Text description of this figure

A coordinate grid with the horizontal axis labeled x running from negative 4 to 5 and the vertical axis labeled y running from negative 1 to 4. Gridlines and number labels sit at every whole number, and the origin is labeled 0. Three filled dots are plotted and labeled with their coordinates: (negative 3, 1), (2, 0) and (4, 3). Nothing is plotted above the input negative 1, the dots are not joined by any segment, and no turning point is marked.
1. Hint 1
  The unknown height must sit strictly between the heights on either side for both neighboring segments to fall.
2. Hint 2
  Translate both falling segments into strict bounds on $k$ , and check whether an integer fits.
Answer

No integer $k$ works; no completed graph or turning point meets all requirements.

Full solution

Falling from $(-3,1)$ to $(-1,k)$ requires $k<1$ .
Falling again from $(-1,k)$ to $(2,0)$ requires $k>0$ .
Together these give
$0<k<1.$
There is no integer strictly between zero and one.

Therefore no integer choice permits the requested graph, and no completed graph or turning point satisfies all the requirements.
Either endpoint value would produce a level segment rather than a decrease.

Answer

No integer $k$ works; no completed graph or turning point meets all requirements.

Key idea

Graph direction requirements may impose strict height bounds with no allowed integer solution.
Problem 6: Mark complete
Problem 7 A threshold record

The graph records a machine’s temperature in degrees Celsius against the time $t$ in minutes during a test. During which time intervals is the temperature above $4$ degrees Celsius, and at what time does the graph change from decreasing to increasing?

A machine's temperature during a six minute test.

Text description of this figure

A graph whose horizontal axis is time in minutes, numbered at every whole number from 0 to 6, and whose vertical axis is temperature in degrees Celsius, numbered at every whole number from 0 to 8, with gridlines at every whole number. The record is three joined straight segments: the temperature rises steadily from 2 degrees at time 0 to 6 degrees at time 2, falls steadily to 2 degrees at time 4, then rises steadily to 6 degrees at time 6. Filled dots mark the two ends and the two corners. No threshold line is drawn, no interval is shaded and no turning time is marked.
1. Hint 1
  Compare the curve with the required temperature level across the whole test.
2. Hint 2
  Find every crossing of that level, then separately inspect the change in direction at the low corner.
Answer

Above $4$ degrees Celsius for $1<t<3$ and $5<t\le6$ minutes; decreasing changes to increasing at $t=4$ minutes.

Full solution

The graph rises from temperature $2$ to $6$ during the first two minutes, meeting temperature $4$ at $t=1$ .
It then falls to $2$ by $t=4$ , crossing $4$ at $t=3$ .
Its final rise reaches $4$ at $t=5$ .

The temperature is strictly above $4$ on $1<t<3$ and $5<t\le6$ minutes.
The equality times are excluded, but the final time is included because the test reaches temperature $6$ there.
The low corner at time $4$ is where decreasing changes to increasing.

Answer

Above $4$ degrees Celsius for $1<t<3$ and $5<t\le6$ minutes; decreasing changes to increasing at $t=4$ minutes.

Key idea

A strict threshold excludes its crossing times while a turning time comes from a change in graph direction.
Problem 7: Mark complete
Problem 8 The endpoint claim

A function has domain $-2\le x\le2$ and both endpoint outputs equal $-1$ . Ana claims its graph has no $x$ -intercepts. Is the claim forced by this information? If not, draw a graph satisfying the endpoint conditions that disproves it and give its $x$ -intercepts.

The two endpoint values, with the grid otherwise empty.

Text description of this figure

A coordinate grid with the horizontal axis labeled x running from negative 3 to 3 and the vertical axis labeled y running from negative 2 to 3. Gridlines and number labels sit at every whole number, and the origin is labeled 0. Exactly two filled dots are plotted, each labeled with its coordinates: (negative 2, negative 1) and (2, negative 1). Nothing joins the two dots, and no other point, segment or curve is drawn.
1. Hint 1
  The endpoint information does not describe all the heights between the endpoints.
2. Hint 2
  Try joining the fixed endpoints to a point above the horizontal axis while preserving one output per input.
Answer

No. One example joins $(-2,-1)$ , $(0,1)$ , and $(2,-1)$ ; its $x$ -intercepts are $(-1,0)$ and $(1,0)$ .

Full solution

Join $(-2,-1)$ to $(0,1)$ and then to $(2,-1)$ with straight segments.
Every input in the specified domain has one output, and both endpoint outputs are correct.

The left segment has rule $y=x+1$ , which is zero at $x=-1$ .
The right segment has rule $y=1-x$ , which is zero at $x=1$ .
Thus this graph has two horizontal-axis intercepts, disproving Ana’s claim.
Other valid counterexamples are accepted.

Answer

No. One example joins $(-2,-1)$ , $(0,1)$ , and $(2,-1)$ ; its $x$ -intercepts are $(-1,0)$ and $(1,0)$ .

Key idea

Endpoint heights alone do not determine whether a function reaches zero between them.
Problem 8: Mark complete
Problem 9 An unbroken climb

A function $m$ has domain $-3\le x\le3$ and range $-2\le y\le4$ , and $m$ increases across its whole domain. Find $m(-3)$ and $m(3)$ , and say how many $x$ -intercepts the graph of $m$ must have. Explain your answers.
1. Hint 1
  The domain is the spread of inputs and the range is the spread of outputs, and increasing keeps those two spreads in step.
2. Hint 2
  Compare $m(x)$ with $m(-3)$ for any other allowed input, and decide which input has to carry the smallest output.
3. Hint 3
  For the intercepts, ask separately whether the output $0$ has to occur at all, and whether it could occur at two different inputs.
Answer

$m(-3)=-2$ and $m(3)=4$ ; the graph has exactly one $x$ -intercept.

Full solution

Take any allowed input $x$ above $-3$ .
Because $m$ increases, the larger input carries the larger output, so
$m(x)>m(-3).$
No output is smaller than $m(-3)$ , which makes it the least output the function produces.
The range names $-2$ as its least value, so $m(-3)=-2$ .

The same comparison at the other end gives $m(x)<m(3)$ for every allowed input below $3$ , so $m(3)$ is the greatest output.
The range names $4$ as its greatest value, so $m(3)=4$ .

The range is the collection of outputs the function actually produces, and $0$ lies between $-2$ and $4$ .
So some allowed input $a$ satisfies $m(a)=0$ , and $(a,0)$ is an $x$ -intercept.

No second input can do the same.
Two inputs with $a<b$ have $m(a)<m(b)$ , so those two outputs differ and cannot both be $0$ .
The graph therefore has exactly one $x$ -intercept.
The line joining $(-3,-2)$ to $(3,4)$ is one graph meeting every condition, with its intercept at $(-1,0)$ , but a different increasing graph puts that intercept somewhere else.

Answer

$m(-3)=-2$ and $m(3)=4$ ; the graph has exactly one $x$ -intercept.

Key idea

An increasing function takes its least and greatest outputs at the ends of its domain, and reaches each output only once.
Problem 9: Mark complete
Problem 10 Two possible drawings

A table gives only $f(0)=0$ , $f(2)=2$ , and $f(4)=0$ . The two drawings both match those records. Jo says the table proves $f(1)=1$ . Does it? Use the drawings to explain.

Two drawings, A and B, through the same three recorded points.

Text description of this figure

Two coordinate grids side by side, labeled A and B. Each has a horizontal axis x running from 0 to 4 and a vertical axis y running from 0 to 3, with gridlines every half unit and number labels at every whole number. In drawing A the graph is two straight segments: a rise from (0, 0) to (2, 2) and a fall from (2, 2) to (4, 0). In drawing B the graph is one smooth curve through the same three points, arching above the straight segments of A between them and flattening at the top near (2, 2). On both grids only the three points (0, 0), (2, 2) and (4, 0) are filled and labeled with their coordinates; no formula is printed, no reading guides are drawn and nothing is marked at the input 1.
1. Hint 1
  A sample table determines its listed points, but it need not determine the graph between them.
2. Hint 2
  Read the height at input one on each drawing and compare what the original records actually specified.
Answer

No; drawing A gives $f(1)=1$ , while drawing B gives about $\tfrac32$ .

Full solution

Drawing A joins the records with straight segments, and its first segment rises one unit per unit of input, so it gives $f(1)=1$ .
Drawing B curves through the same three records, and above input $1$ it meets the half-unit gridline at height $3/2$ .

Both drawings have one output per input and match all three listed table entries.
Because they disagree at an unlisted input, the table alone does not determine $f(1)$ .
Jo’s value would need an additional statement about the graph between the recorded points.

Answer

No; drawing A gives $f(1)=1$ , while drawing B gives about $\tfrac32$ .

Key idea

Matching a finite sample does not force two functions to agree at unlisted inputs.
Problem 10: Mark complete

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