SAT Nonlinear Functions Question 165 | Free SAT Prep

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Question 165·Hard·Nonlinear Functions

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The cubic polynomial $p$ is defined by

p(x)=x^3-ax^2+9ax-27,

where $a$ is a constant. In the $xy$ -plane, the graph of $y=p(x)$ has a local minimum at $x=3$ . What is the value of $a$ ?

Your answer

For a polynomial with a local maximum or minimum at a specified input, compare $p(x)$ with the value at that input. The difference $p(x)-p(r)$ has a factor of $x-r$ . At a turning point, that factor must occur twice, so set the remaining factor equal to $0$ at $x=r$ . This approach uses polynomial structure rather than calculus.

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Compare nearby outputs to $p(3)$

First calculate $p(3)$ . Then consider the expression $p(x)-p(3)$ , which equals $0$ when $x=3$ .

Factor using the known zero

Since $p(x)-p(3)=0$ at $x=3$ , factor $p(x)-p(3)$ using $x-3$ as a factor.

Use the behavior at a local minimum

At a local minimum, the graph does not pass through the point at $x=3$ ; it turns there. Because $x-3$ changes sign at $3$ , what must happen to the other factor?

Desmos Guide

Graph the condition on $a$

After factoring $p(x)-p(3)$ by hand, substitute $x=3$ into the remaining factor. Enter y=27+3x in Desmos. The $x$ -coordinate of its intercept gives the value that $a$ must have.

Verify the local minimum

Substitute the parameter value into the original polynomial and graph it in Desmos. Check that the graph turns upward at $x=3$ , confirming a local minimum there.

Step-by-step Explanation

Find the polynomial's value at $x=3$

Substitute $x=3$ into the polynomial:

\begin{aligned} p(3)&=3^3-a(3^2)+9a(3)-27\\ &=27-9a+27a-27\\ &=18a. \end{aligned}

Factor the difference from the value at $x=3$

Because $p(3)=18a$ , subtracting this value gives

\begin{aligned} p(x)-p(3) &=x^3-ax^2+9ax-27-18a\\ &=(x-3)\left(x^2+(3-a)x+9+6a\right). \end{aligned}

The factor $x-3$ shows that this difference is $0$ when $x=3$ .

Use the local-minimum condition

At a local minimum, $p(x)-p(3)$ must not change from positive to negative, or from negative to positive, as $x$ passes through $3$ . Since the factor $x-3$ changes sign at $3$ , the other factor must also equal $0$ at $x=3$ .

\begin{aligned} 3^2+(3-a)(3)+9+6a&=0\\ 27+3a&=0. \end{aligned}

Therefore,

a=-9.

Confirm that the extremum is a minimum

With $a=-9$ ,

p(x)-p(3)=(x-3)^2(x+15).

For values of $x$ near $3$ , $x+15$ is positive, and $(x-3)^2$ is nonnegative. Thus $p(x)\ge p(3)$ near $x=3$ , confirming that the point is a local minimum.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation