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Polynomials and Their 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 One expression, one test

Determine whether $f(x)=\pi x^4-8x+\frac19$ is a polynomial. If it is, give its degree and leading coefficient; if it is not, name the specific requirement it fails.
1. Hint 1
  A polynomial restricts what happens to the variable $x$ in each term; a coefficient may be any real number.
2. Hint 2
  Check every term for $x$ divided into, rooted, or placed in an exponent, and note each exponent on $x$ .
Answer

Yes, it is a polynomial; degree $4$ , leading coefficient $\pi$ .

Full solution

Each term of $f$ has a whole-number exponent on $x$ : $4$ in $\pi x^4$ , $1$ in $-8x$ , and $0$ in the constant $\frac19$ , and $x$ is never divided into, rooted, or placed in an exponent.
$f(x)=\pi x^4-8x+\frac19$
So $f$ is a polynomial even though one coefficient, $\pi$ , is irrational, since the restriction falls only on the variable's own exponent.
The terms are already in descending order, so the leading term is $\pi x^4$ , giving degree $4$ and leading coefficient $\pi$ .

Answer

Yes, it is a polynomial; degree $4$ , leading coefficient $\pi$ .

Key idea

A coefficient may be any real number, including an irrational one; only the variable's own exponent decides whether an expression is a polynomial.
Problem 1: Mark complete
Problem 2 A product label

A nonzero polynomial $A(x)$ has leading term $-3x^4$ . The product $A(x)B(x)$ has leading term $12x^9$ . Find the leading term of $B(x)$ .
1. Hint 1
  The highest powers in a product come from its factors' leading terms.
2. Hint 2
  The degrees add, while the leading coefficients multiply.
Answer

$-4x^5$ .

Full solution

Write the leading term of $B$ as $bx^n$ , with $b\ne0$ .
Then the leading term of the product is $-3b x^{n+4}$ .
Matching the given power and coefficient gives
$\begin{aligned} n+4&=9 \\ -3b&=12. \end{aligned}$
Thus $n=5$ and $b=-4$ , so the required term is $-4x^5$ .
Multiplying it by $-3x^4$ checks the given product term.

Answer

$-4x^5$ .

Key idea

A product's leading term determines one factor's leading term when the other is known.
Problem 2: Mark complete
Problem 3 A difference of readings

For $f(x)=3x^4-2x^3+6x-5$ , find $f(1)-f(0)$ .
1. Hint 1
  At zero only the constant remains; at one every power equals one.
2. Hint 2
  The constant cancels in the requested difference.
Answer

$7$ .

Full solution

The evaluations are $f(1)=2$ and $f(0)=-5$ .
$f(1)-f(0)=2-(-5)$
This equals $7$ , also the sum $3-2+6$ of the nonconstant coefficients.

Answer

$7$ .

Key idea

The difference between the values at one and zero equals the sum of the nonconstant coefficients.
Problem 3: Mark complete
Problem 4 A family of rules

Let $F_k(x)=(k-1)x^4+(k+2)x^3-5$ . Find the real $k$ for which $F_k$ has degree $3$ , then state its leading coefficient, constant term, and far-end behavior.
1. Hint 1
  The fourth-degree term must vanish while the cubic term remains.
2. Hint 2
  After choosing $k$ , read the sign and parity of the surviving leading term.
Answer

$k=1$ ; leading coefficient $3$ , constant $-5$ ; far left falls and far right rises.

Full solution

Degree $3$ requires $k-1=0$ , so $k=1$ .
At this value $k+2=3\ne0$ , and
$F_1(x)=3x^3-5$
The positive cubic leading term makes the far left fall and the far right rise.
The constant is $-5$ .

Answer

$k=1$ ; leading coefficient $3$ , constant $-5$ ; far left falls and far right rises.

Key idea

A parameter can change the degree when it cancels the highest-power coefficient.
Problem 4: Mark complete
Problem 5 A box model

A box has side lengths $x$ , $x+1$ , and $2x+3$ meters for real $x>0$ . Write its volume polynomial in standard form. For that polynomial extended to all real inputs, give the degree, leading term, and far-end behavior.
1. Hint 1
  Volume multiplies the three lengths.
2. Hint 2
  Expand or combine the leading terms, then distinguish the physical domain from the polynomial domain.
Answer

$V(x)=2x^3+5x^2+3x$ cubic meters; degree $3$ , leading term $2x^3$ ; far left falls, far right rises.

Full solution

First
$(x+1)(2x+3)=2x^2+5x+3.$
Multiplying by $x$ gives
$V(x)=2x^3+5x^2+3x$
Its degree is $3$ and its leading term is $2x^3$ .
As a polynomial on all real inputs its far left falls and far right rises, although negative inputs do not describe the stated physical box.

Answer

$V(x)=2x^3+5x^2+3x$ cubic meters; degree $3$ , leading term $2x^3$ ; far left falls, far right rises.

Key idea

A product model has degree equal to the sum of factor degrees, while its physical domain may be smaller.
Problem 5: Mark complete
Problem 6 A visible curve

The figure shows the complete zero and turning-point pattern of a degree- $4$ polynomial. Read its distinct real zeros and count its turning points. Explain whether either count exceeds the degree bounds.

The graph of a degree- $4$ polynomial.

Text description of this figure

A grid with the x-axis from -1.5 to 1.5, ticked every 0.5, and the y-axis from -0.5 to 2, ticked every 0.25. A single smooth curve meets the horizontal axis at exactly three points, x equals -1, x equals 0, and x equals 1. On each side of x equals 0 the curve dips below the axis into a valley, and at x equals 0 it rises only as far back as the axis, at a peak between the two valleys, before turning back down into the second valley. Near the top edge of the frame, both ends of the curve are marked with an arrow showing that they continue rising beyond what is drawn. No equation, coordinates, or turning points are labeled.
1. Hint 1
  A real zero is an input where the curve meets the horizontal axis.
2. Hint 2
  Degree $4$ permits at most four distinct real zeros and three turning points.
Answer

Zeros $-1,0,1$ ; three turning points; neither count exceeds its bound.

Full solution

The curve meets the axis at $-1$ , $0$ , and $1$ , so there are three distinct real zeros.
It turns in the two valleys and at the central peak, giving three turning points.
The bounds are
$3\le4$
for real zeros and
$3\le4-1$
for turning points.
Both counts are allowed.

Answer

Zeros $-1,0,1$ ; three turning points; neither count exceeds its bound.

Key idea

Degree gives ceilings on distinct zeros and turning points, not a requirement that both ceilings be reached.
Problem 6: Mark complete
Problem 7 Three coefficient clues

A cubic polynomial has leading coefficient $2$ , no $x^2$ term, $f(0)=5$ , and $f(1)=-1$ . Find $f$ in standard form and give the largest possible number of its real zeros and turning points.
1. Hint 1
  The clues fix the cubic and constant coefficients, leaving one unknown coefficient.
2. Hint 2
  Evaluate the partially known polynomial at one to determine that coefficient.
Answer

$f(x)=2x^3-8x+5$ ; at most three real zeros and two turning points.

Full solution

The form is
$f(x)=2x^3+bx+5.$
Since $f(1)=-1$ ,
$2+b+5=-1$
Thus $b=-8$ and
$f(x)=2x^3-8x+5.$
Its degree $3$ permits at most three real zeros and two turning points.
The values $f(0)=5$ and $f(1)=-1$ check the coefficient clues.

Answer

$f(x)=2x^3-8x+5$ ; at most three real zeros and two turning points.

Key idea

Values at zero and one can recover missing polynomial coefficients efficiently.
Problem 7: Mark complete
Problem 8 A statement about two rules

Two monic polynomials each have degree $5$ . A student says their difference is either zero or has degree at most $4$ . Is that correct? Justify your answer.
1. Hint 1
  Monic means each leading coefficient is one.
2. Hint 2
  Inspect what happens to the two $x^5$ terms when subtracting.
Answer

Correct; the difference is zero or has degree at most $4$ .

Full solution

Write the polynomials as $x^5+A(x)$ and $x^5+B(x)$ , where $A$ and $B$ have degree at most $4$ or are zero.
Their difference is
$(x^5+A)-(x^5+B)=A-B$
Thus no fifth-degree term survives.
The result is zero or has degree at most $4$ .

Answer

Correct; the difference is zero or has degree at most $4$ .

Key idea

Equal leading terms cancel in subtraction, lowering the possible degree.
Problem 8: Mark complete
Problem 9 A constant graph

A student applies the rule of at most $n-1$ turning points to $f(x)=-7$ , obtaining at most $-1$ turning points. Explain what went wrong and give the actual numbers of zeros and turning points.
1. Hint 1
  The turning-point formula is stated for nonconstant polynomials.
2. Hint 2
  Picture the horizontal line at height $-7$ .
Answer

Zero real zeros and zero turning points; the $n-1$ bound requires a nonconstant polynomial.

Full solution

The function has degree zero, but the turning-point bound $n-1$ applies only when $n\ge1$ .
Its graph is the horizontal line $y=-7$ .
It never meets $y=0$ and never switches from rising to falling, so it has no real zeros and no turning points.

Answer

Zero real zeros and zero turning points; the $n-1$ bound requires a nonconstant polynomial.

Key idea

Constant polynomials need their own turning-point statement.
Problem 9: Mark complete
Problem 10 A proposed conclusion

A polynomial $p$ has even positive degree and $p(0)<0$ . A student concludes that it must have a positive real zero. Is the conclusion justified? Give a counterexample if it is false.
1. Hint 1
  Even degree makes the ends agree but does not specify whether they rise or fall.
2. Hint 2
  Consider a downward-opening quadratic whose values are all negative.
Answer

False; $p(x)=-x^2-1$ is a counterexample.

Full solution

Take
$p(x)=-x^2-1$
It has even degree $2$ and $p(0)=-1<0$ .
Since $x^2\ge0$ for real $x$ , $p(x)\le-1$ , so it has no real zero at all.
The missing information is whether the leading coefficient is positive.

Answer

False; $p(x)=-x^2-1$ is a counterexample.

Key idea

Parity determines whether the ends agree, while the leading coefficient determines their direction.
Problem 10: Mark complete

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