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Graphing Polynomial 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 Reading a sign

The figure shows a polynomial curve, and every x-intercept is visible. On which open intervals is its value negative?

The graph of the polynomial.

Text description of this figure

A grid with the x-axis from -5.5 to 3.5, ticked at every integer, and the y-axis from -18 to 18, ticked every 6. A smooth curve rises off the top of the frame on the left, crosses the x-axis at x equals -4, dips down, rises back up crossing at x equals -1, reaches a peak, then descends, crossing at x equals 2 and continuing off the bottom of the frame on the right. Arrows at both ends show the curve continuing beyond the frame. No equation or sign labels are shown.
1. Hint 1
  Negative values appear below the horizontal axis.
2. Hint 2
  Read the three crossing locations and separate the intervals they create.
Answer

$-4<x<-1$ and $x>2$ .

Full solution

The curve crosses at $-4$ , $-1$ , and $2$ .
It is below the axis between the first two and to the right of the last one.
Those regions are
$-4<x<-1$
and $x>2$ .
The zeros themselves are excluded because the question asks for negative values.

Answer

$-4<x<-1$ and $x>2$ .

Key idea

The intervals below the horizontal axis are exactly where a graph has negative values.
Problem 1: Mark complete
Problem 2 An added factor

A real polynomial $p$ has exactly three distinct real zeros. How many x-intercepts does the graph of $q(x)=(x^2+2x+10)^3p(x)$ have?
1. Hint 1
  A new intercept requires the extra factor to vanish at a real input.
2. Hint 2
  For real $x$ , compare $x^2+2x+10$ with zero.
Answer

$3$ x-intercepts.

Full solution

For every real $x$ , $x^2+2x+10=(x+1)^2+9$ , which is at least $9$ , so it is positive.
Its cube is also positive.
Thus $q(x)=0$ exactly at the real zeros of $p$ , so the three distinct intercepts are unchanged.
The added factor has nonreal roots, which add no x-intercepts.

Answer

$3$ x-intercepts.

Key idea

A factor that is positive at every real input adds no x-intercepts.
Problem 2: Mark complete
Problem 3 An interval report

A polynomial $p$ has no real zero in $-1<x<5$ , and $p(2)=-7$ . Determine the sign of $p(4)$ .
1. Hint 1
  A polynomial is continuous and cannot change sign without reaching zero.
2. Hint 2
  The known input and the requested input are inside the same interval without zeros.
Answer

$p(4)<0$ .

Full solution

The value at $2$ is negative.
If $p(4)$ were positive, the unbroken polynomial graph would have to pass through zero between $2$ and $4$ .
If $p(4)=0$ , it would itself be a forbidden zero.
Therefore
$p(4)<0$
No exact evaluation is needed.

Answer

$p(4)<0$ .

Key idea

One sign test determines every value's sign within an interval containing no real zero.
Problem 3: Mark complete
Problem 4 A quartic from its graph

The figure shows every x-intercept of a degree $4$ polynomial and the point $(1,50)$ . Find the polynomial in factored form.

The graph of the polynomial, with one labeled point.

Text description of this figure

A grid with the x-axis from -5 to 7, ticked at every integer, and the y-axis from -180 to 60, ticked every 20. A smooth curve rises off the top of the frame on the left, comes down and crosses the x-axis at x equals -4, dips, rises back up crossing at x equals 0, rises to a peak, then descends, crossing at x equals 3, dips again, crosses at x equals 6, and rises off the top of the frame on the right. A solid point is marked at (1, 50), with that coordinate label. No equation is shown.
1. Hint 1
  The four crossings account for the whole degree.
2. Hint 2
  Use the additional plotted point to determine the remaining constant.
Answer

$p(x)=x(x+4)(x-3)(x-6)$ .

Full solution

The crossings at $-4,0,3,6$ must all be simple because four distinct roots already use degree $4$ .
Thus
$p(x)=ax(x+4)(x-3)(x-6)$

At the labeled point,
$p(1)=a(1)(5)(-2)(-5)$ $50=50a$ $a=1$
The resulting polynomial has the four observed roots, positive leading coefficient, and $p(1)=50$ , as required.

Answer

$p(x)=x(x+4)(x-3)(x-6)$ .

Key idea

A known degree and enough intercepts can fix all multiplicities before an extra point determines the scale.
Problem 4: Mark complete
Problem 5 Comparing two curves

Let $p(x)=x(x-2)^2$ and $q(x)=(x^2+3x+5)p(x)$ . Compare their degrees, x-intercepts, crossing or touching behavior, and signs for real inputs.
1. Hint 1
  The added factor contributes degree but is positive for every real input.
2. Hint 2
  Read multiplicities from $p$ , then check whether the extra factor vanishes at either zero.
Answer

Degrees $3$ and $5$ ; both cross at $0$ and touch at $2$ ; both negative for $x<0$ , positive for $x>0$ except at $2$ .

Full solution

The degrees are $3$ for $p$ and $5$ for $q$ .
The quadratic $x^2+3x+5$ has discriminant $9-20=-11$ , which is negative, so it is positive for every real $x$ .
Thus $q$ has the same real zeros and multiplicities as $p$ : $0$ is simple and $2$ has multiplicity $2$ .
Both cross at $0$ and touch at $2$ .

Away from $x=2$ , $(x-2)^2$ is positive, so the sign of $p$ is the sign of $x$ .
The positive extra factor preserves that sign for $q$ .
For example,
$p(-1)=-9$ $p(1)=1$ $p(3)=3$
These signs cover all three intervals.

Answer

Degrees $3$ and $5$ ; both cross at $0$ and touch at $2$ ; both negative for $x<0$ , positive for $x>0$ except at $2$ .

Key idea

A positive polynomial factor can raise the degree without changing real zeros or signs.
Problem 5: Mark complete
Problem 6 Sketching a factored expression

For $p(x)=-x^2(x+2)(x-1)$ , sketch the graph using its intercepts, signs, and end behavior. Give the sign on each interval between real zeros and state which intercepts are crossings or touches. Do not assign coordinates to uncomputed turning points.
1. Hint 1
  Read the degree and multiplicities before choosing test inputs.
2. Hint 2
  Use one input in each interval cut out by $-2$ , $0$ , and $1$ .
Answer

Both ends down; crosses at $-2,1$ , touches at $0$ ; signs $-,+,+,-$ on the four intervals in increasing order.

Full solution

The leading term is $-x^4$ , so both ends go down.
The zeros $-2$ and $1$ are simple, while $0$ has multiplicity $2$ .
Thus $-2$ and $1$ are crossings and $0$ is a touch.

Four test values give
$p(-3)=-36$ $p(-1)=2$ $p(\tfrac12)=\tfrac5{16}$ $p(2)=-16$

The graph is below the axis before $-2$ , above it on $(-2,0)$ and $(0,1)$ , and below it after $1$ .
It touches from above at $0$ .
These features determine the requested sketch; no exact off-axis turning-point location has been established.

The completed sketch of $p$ .

Answer

Both ends down; crosses at $-2,1$ , touches at $0$ ; signs $-,+,+,-$ on the four intervals in increasing order.

Key idea

An honest polynomial sketch combines end behavior, root multiplicities, and interval signs.
Problem 6: Mark complete
Problem 7 Labels on a plotted curve

The graph shows $p(x)=x^2(x+1)(x-3)$ and a marked turning point A at the origin. A student labels A as a local maximum and labels the minimum function value as $-16$ because that is the bottom of the plotting window. Which label is justified by the stated evidence? Explain without calculus.

The graph of $p$ , with turning point A marked.

Text description of this figure

A grid with the x-axis from -2 to 4, ticked at every integer, and the y-axis from -16 to 20, ticked every 4. A smooth curve rises off the top of the frame on the left, descends and crosses the x-axis at x equals -1, dips down and touches the axis from below at the origin, marked A, dips to a minimum, then rises and crosses the axis at x equals 3, continuing off the top of the frame on the right. No other point or turning-point coordinate is labeled.
1. Hint 1
  The sign on either side of an even-multiplicity zero determines its local type.
2. Hint 2
  A viewing-window boundary is a plotting choice, not a computed function extremum.
Answer

Only A as a local maximum is justified; the minimum-value label is not.

Full solution

Near $0$ , $x^2$ is positive except at $0$ , $x+1$ is positive, and $x-3$ is negative.
Thus the curve is below the axis on both sides of $0$ , proving that A is a local maximum.

The value $-16$ was chosen as the plotting-window boundary.
No calculation establishes that $p$ attains that value or that it is the least value of $p$ .
A window label is not evidence for an exact extremum.

Answer

Only A as a local maximum is justified; the minimum-value label is not.

Key idea

A plotting-window boundary does not determine a polynomial curve's exact maximum or minimum.
Problem 7: Mark complete
Problem 8 Two positive readings

A student knows only that a real polynomial satisfies $p(-3)>0$ and $p(3)>0$ . The student concludes that it has no real zero between those inputs. Is the conclusion justified? Give a counterexample if needed.
1. Hint 1
  Positive endpoint values do not control whether the curve dips to the axis in between.
2. Hint 2
  A polynomial with a repeated root can be positive at both endpoints.
Answer

No; for example, $p(x)=x^2$ has a zero at $0$ and both endpoint values equal $9$ .

Full solution

Take $p(x)=x^2$ .
Then
$p(-3)=9$ $p(3)=9$
but $p(0)=0$ , so the claimed conclusion fails.

A polynomial must have a zero to change sign, but a zero need not produce a sign change.
Here the even-multiplicity zero merely touches the axis.

Answer

No; for example, $p(x)=x^2$ has a zero at $0$ and both endpoint values equal $9$ .

Key idea

Equal signs at two endpoints do not rule out real zeros between them.
Problem 8: Mark complete
Problem 9 A curve without crossings

A nonconstant polynomial $p$ with real coefficients has no x-intercepts, and $p(0)<0$ . A student claims that $p(x)<0$ for every real $x$ . Is the claim true? Explain.
1. Hint 1
  There is no real zero anywhere to allow a sign change.
2. Hint 2
  Compare a hypothetical positive value with the known negative value at $0$ .
Answer

True; $p(x)<0$ for every real $x$ .

Full solution

A zero value would itself be an x-intercept.
If some input had a positive value, continuity between that input and $0$ would force an x-intercept as well.
Both contradict the premise.
Hence
$p(x)<0$
for every real $x$ .

Answer

True; $p(x)<0$ for every real $x$ .

Key idea

A polynomial with no real zeros keeps one sign on the entire real line.
Problem 9: Mark complete
Problem 10 What a sketch determines

The figure shows all x-intercepts and both end directions of a nonzero real polynomial. Find its least possible degree and the sign of its leading coefficient. Does the sketch prove the exact multiplicities or the exact locations of its nonreal roots? Explain.

A schematic view of the polynomial.

Text description of this figure

A grid with the x-axis from -5 to 1 and the y-axis from -6 to 2, both with unit ticks. A smooth curve rises from the lower left, touches the x-axis from below at x equals -3 without crossing it, dips down, rises back up to touch the axis from below at x equals 0, then falls off the bottom of the frame on the right. Arrows at both ends show the curve continuing downward beyond the frame. No equation, coordinates, or multiplicity is labeled.
1. Hint 1
  A crossing needs at least one factor and a touch needs at least two.
2. Hint 2
  Extra even multiplicities or nonreal conjugate pairs can preserve the displayed real-axis behavior.
Answer

Least degree $4$ ; negative leading coefficient; exact multiplicities and nonreal root locations are not established.

Full solution

The two displayed intercepts are touches, each with even multiplicity at least $2$ .
Thus
$n\ge2+2$ $n\ge4$
Both ends fall, so the degree is even and the leading coefficient is negative.

The curve shows even multiplicity at each root, not an exact exponent.
Higher even multiplicities or extra factors with no real roots can preserve the same real-axis features.
No nonreal root is a point on this real graph.

Answer

Least degree $4$ ; negative leading coefficient; exact multiplicities and nonreal root locations are not established.

Key idea

A schematic curve supports multiplicity parity and a minimum degree, but not a complete complex root list.
Problem 10: Mark complete

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