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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 Four allowed points

The grid is provided for a function with domain $\{-2,0,1,3\}$ and rule $f(x)=2-\lvert x\rvert$ . Draw its entire graph.

A blank coordinate grid for the graph.

Text description of this figure

A grid with the horizontal axis labeled x running from -3 to 4 and the vertical axis labeled y running from -2 to 3, gridlines and number labels at every whole number, and the origin labeled 0. No point or curve is drawn.
1. Hint 1
  The graph has one point for each declared input.
2. Hint 2
  Evaluate the rule at the four inputs; the domain contains no values between them.
Answer

The four points $(-2,0)$ , $(0,2)$ , $(1,1)$ , and $(3,-1)$ , with no connecting lines.

Full solution

Compute the outputs.
$\begin{aligned}f(-2)&=0\\f(0)&=2\\f(1)&=1\\f(3)&=-1\end{aligned}$

Plot these four ordered pairs.
Do not connect them: any connecting segment would add inputs absent from the declared domain.

Answer

The four points $(-2,0)$ , $(0,2)$ , $(1,1)$ , and $(3,-1)$ , with no connecting lines.

Key idea

A graph contains exactly the pairs allowed by both the rule and its declared domain.
Problem 1: Mark complete
Problem 2 A product of factors

Find all real zeros of $f(x)=(x^2+4)(2x-1)(x+2)^2$ .
1. Hint 1
  A product is zero when one of its factors is zero.
2. Hint 2
  One factor is positive for every real input, while the other two can vanish.
Answer

$x=-2$ and $x=\frac12$ .

Full solution

The factor $x^2+4$ is at least $4$ , so it has no real zero.
The other factors give
$\begin{aligned}2x-1&=0\\x&=\frac12\end{aligned}$
and $(x+2)^2=0$ , so $x=-2$ .

Each reported input makes a factor vanish, and there are no other ways for the product to be zero.

Answer

$x=-2$ and $x=\frac12$ .

Key idea

A factor that is strictly positive introduces no real zero into a product.
Problem 2: Mark complete
Problem 3 A magnitude ratio

On the domain $x\ne0$ , classify $f(x)=\frac{\lvert x\rvert}{x}$ as even, odd, or neither by examining $f(-x)$ .
1. Hint 1
  The domain must include the opposite of every allowed input.
2. Hint 2
  Changing $x$ to $-x$ leaves its magnitude unchanged and changes the denominator sign.
Answer

Odd, and not even.

Full solution

The domain is symmetric about zero.
For an allowed input,
$\begin{aligned}f(-x)&=\frac{\lvert x\rvert}{-x}\\&=-f(x)\end{aligned}$

Thus the function is odd.
It is not even because $f(1)=1$ and $f(-1)=-1$ .

Answer

Odd, and not even.

Key idea

Test symmetry by replacing the full input, including its appearances inside other operations.
Problem 3: Mark complete
Problem 4 A table that does not pin down the graph

Let $f(x)=x^2$ for all real $x$ . Give a rule $g(x)$ , different from $f(x)$ for at least one input, that agrees with $f$ at $x=0$ , $x=2$ , and $x=4$ . Then explain why matching $f$ at these three inputs cannot force $g=f$ everywhere.
1. Hint 1
  A factor built from the three sample points is zero at each one, so adding any multiple of it to $f$ leaves those three outputs unchanged.
2. Hint 2
  Multiply that factor by any nonzero constant and add the result to $f$ .
Answer

For example, $g(x)=x^2+x(x-2)(x-4)$ ; it agrees with $f$ at $0,2,4$ , but $g(1)=4\ne1=f(1)$ .

Full solution

Let $p(x)=x(x-2)(x-4)$ , which is $0$ at each sample point since one factor vanishes there.
Adding any multiple of $p$ to $f$ changes nothing at $0,2,4$ :
$\begin{aligned}g(0)&=f(0)+p(0)=0+0=0,\\g(2)&=f(2)+p(2)=4+0=4,\\g(4)&=f(4)+p(4)=16+0=16.\end{aligned}$

Away from the sample points $p$ is nonzero, so $g$ and $f$ can differ there: $g(1)=1+(1)(-1)(-3)=4$ , while $f(1)=1$ .
A finite table of three matching outputs is satisfied by infinitely many different choices of the added constant, so it cannot pin down a single function.

Answer

For example, $g(x)=x^2+x(x-2)(x-4)$ ; it agrees with $f$ at $0,2,4$ , but $g(1)=4\ne1=f(1)$ .

Key idea

A finite table of outputs is satisfied by infinitely many different functions, so only a rule proven from its structure determines a graph.
Problem 4: Mark complete
Problem 5 A replaced point

A function on $\mathbb R$ follows $f(x)=2x+1$ for $x\ne-1$ , but has $f(-1)=-2$ . Draw the graph on the provided grid, and give the point removed from the line and the full range.

A blank coordinate grid for the graph.

Text description of this figure

A grid with the horizontal axis labeled x running from -4 to 3 and the vertical axis labeled y running from -5 to 5, gridlines and number labels at every whole number, and the origin labeled 0. No point or curve is drawn.
1. Hint 1
  The exceptional input replaces one point of the line with another point.
2. Hint 2
  Find the usual line height at $x=-1$ , then check whether its replacement height occurs elsewhere.
Answer

Line $y=2x+1$ with an open point at $(-1,-1)$ and filled point at $(-1,-2)$ ; range $\mathbb R\setminus\{-1\}$ .

Full solution

The usual line would have height
$2(-1)+1=-1$
at input $-1$ , but this point is removed.
Place a filled point at the declared value $(-1,-2)$ instead.

Every other line height remains, and height $-2$ already occurs on the line at
$x=-\frac32.$
The line produces height $-1$ only at the excluded line input $-1$ , so the range is every real value except $-1$ .

Answer

Line $y=2x+1$ with an open point at $(-1,-1)$ and filled point at $(-1,-2)$ ; range $\mathbb R\setminus\{-1\}$ .

Key idea

A special value in a piecewise rule can replace a point without removing that input from the domain.
Problem 5: Mark complete
Problem 6 Far from the center

Let $f(x)=-2x^5+x^3$ for real $x$ . Describe both ends of its graph. Justify your conclusion by taking out the highest power and bounding the remaining factor when $\lvert x\rvert\ge1$ .
1. Hint 1
  A highest-power factor separates growth from a smaller correction.
2. Hint 2
  For $\lvert x\rvert\ge1$ , the correction $1/x^2$ lies between $0$ and $1$ .
Answer

The left end rises without bound; the right end falls without bound.

Full solution

For $x\ne0$ ,
$f(x)=x^5\left(-2+\frac1{x^2}\right).$
When $\lvert x\rvert\ge1$ , the factor in parentheses is at most $-1$ and greater than $-2$ .

Thus $f(x)$ has the opposite sign to $x^5$ and magnitude at least $\lvert x\rvert^5$ .
This magnitude grows beyond any bound: the far-left values are positive and the far-right values are negative.

Answer

The left end rises without bound; the right end falls without bound.

Key idea

A bounded nonzero correction factor preserves the growth and fixes the sign supplied by the highest power.
Problem 6: Mark complete
Problem 7 A curve with a coefficient

The graph shows $f(x)=a(x+1)^2(x-2)$ for a nonzero real constant $a$ . Read enough information to determine $a$ , then explain whether the graph crosses or touches the horizontal axis at each zero.

The graph of the function, clipped at the edges of the grid.

Text description of this figure

A grid with the horizontal axis labeled x running from -3 to 3 and the vertical axis labeled y running from -5 to 6, gridlines and number labels at every whole number, and the origin labeled 0. A single smooth curve enters through the top edge of the grid a little to the left of x = -2, descends to a local low point near (-1, 0), rises to a local high point near (1, 4), then descends again and leaves through the bottom edge a little to the right of x = 2. The curve passes through (0, 2). No point is marked, no formula is written, and the curve is not labeled.
1. Hint 1
  One readable nonzero output determines the remaining constant.
2. Hint 2
  Use the height at input $0$ , then examine the powers of the two factors.
Answer

$a=-1$ ; touches at $x=-1$ and crosses at $x=2$ .

Full solution

The graph has height $2$ at input $0$ .
Substitute this point into the rule.
$\begin{aligned}2&=a(1)^2(-2)\\a&=-1\end{aligned}$

The zero at $-1$ has even multiplicity $2$ , so the sign does not change there and the graph touches.
The zero at $2$ has odd multiplicity $1$ , so the sign changes and the graph crosses.
The constant $-1$ reverses the overall signs but not these crossing decisions.

Answer

$a=-1$ ; touches at $x=-1$ and crosses at $x=2$ .

Key idea

A point can determine an unknown coefficient while factor powers determine local crossing behavior.
Problem 7: Mark complete
Problem 8 An unrecorded output

A function has declared domain $\{-1,0,1\}$ and codomain $\{0,1\}$ . The only recorded outputs are $f(-1)=0$ and $f(1)=1$ . A student says there are exactly two possible complete graphs. Is that correct? Explain.
1. Hint 1
  The recorded values already fix two of the three graph points.
2. Hint 2
  Only one input remains, and its output must lie in the two-element codomain.
Answer

Yes; the remaining point is either $(0,0)$ or $(0,1)$ .

Full solution

Every graph must contain $(-1,0)$ and $(1,1)$ .
The only undecided input is $0$ , and its output must be $0$ or $1$ .

Thus there are exactly two possible remaining pairs, $(0,0)$ and $(0,1)$ .
Each completes a valid function, and the declared domain has no other inputs to choose.

Answer

Yes; the remaining point is either $(0,0)$ or $(0,1)$ .

Key idea

A finite table can determine nearly all of a function when its declared domain is also finite.
Problem 8: Mark complete
Problem 9 Two positive samples

Let $p(x)=(x-2)(x-5)$ for real $x$ . A student finds $p(0)>0$ and $p(6)>0$ and concludes the graph stays above the horizontal axis for every $0<x<6$ . Assess the conclusion using the factors.
1. Hint 1
  The sampled endpoints do not determine the signs between them.
2. Hint 2
  Consider the interval between the two roots, where the factors have opposite signs.
Answer

False; $p(x)<0$ on $2<x<5$ , with zeros at $2$ and $5$ .

Full solution

For $2<x<5$ , $x-2>0$ and $x-5<0$ , so their product is negative.
For example,
$p(3)=-2.$

The graph is above the axis on $(0,2)$ and $(5,6)$ , meets it at $2$ and $5$ , and is below it between them.
Two positive sample values did not establish the intervening behavior.

Answer

False; $p(x)<0$ on $2<x<5$ , with zeros at $2$ and $5$ .

Key idea

Factor signs can reveal behavior that a small collection of samples misses.
Problem 9: Mark complete
Problem 10 A restricted picture

The formula $f(x)=x^2+2$ is assigned the domain $[-2,3]$ . A student says its graph is symmetric across the vertical axis because replacing $x$ by $-x$ leaves the formula unchanged. Is the conclusion correct for this declared graph?
1. Hint 1
  A graph includes its domain, not just its formula.
2. Hint 2
  Check whether the domain contains the opposite of its largest input.
Answer

No; $(3,11)$ is present but its reflection $(-3,11)$ is absent.

Full solution

The formula does satisfy
$(-x)^2+2=x^2+2.$

However, $3$ is allowed and $-3$ is not.
Thus the graph contains $(3,11)$ but not $(-3,11)$ , so reflection does not preserve this declared graph.
The symmetric-domain requirement fails.

Answer

No; $(3,11)$ is present but its reflection $(-3,11)$ is absent.

Key idea

An unchanged formula does not establish graph symmetry when the declared domain lacks that symmetry.
Problem 10: Mark complete

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