Question 130·Hard·Math

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The graph of a polynomial function $f$ is shown.

The polynomial $f$ has the least possible degree and real coefficients.

Which choice is the value of $f(3)$ ?

When a graph shows intercepts, build the polynomial in factored form. A key harder detail is multiplicity: if the graph crosses the x-axis at a root, the factor appears once; if it only touches and turns around, that factor is squared (or another even power, but “least possible degree” means square it). After writing $f(x)=a\cdot(\text{factors})$ , use one labeled point on the graph to solve for $a$ , and only then plug in the requested x-value.

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Use the x-intercepts

The x-intercepts tell you factors of $f(x)$ . Each intercept $x=r$ gives a factor $(x-r)$ .

Watch what happens at $x=-2$

At one intercept, the graph touches the x-axis and turns back up instead of crossing. That tells you the corresponding factor is squared.

Use the labeled point

After you build the factored form with an unknown constant multiplier, plug in the labeled point on the graph to find that constant.

Desmos Guide

Enter a factored form with a slider

Type

$y=a(x+2)^2(x-1)(x-4)$

and let Desmos create the slider $a$ .

Use the labeled point to set $a$

Plot the point $(0,2)$ . Adjust the slider $a$ until the graph passes through that point.

Find the value at $x=3$

With that $a$ , either evaluate $a(3+2)^2(3-1)(3-4)$ in a new expression line or make a table for the function and look at the y-value when $x=3$ . Then match the result to the answer choices.

Step-by-step Explanation

Write the polynomial from its x-intercepts

From the graph, the x-intercepts are at $x=-2$ , $x=1$ , and $x=4$ .

At $x=-2$ , the graph touches the x-axis and turns around, so $(x+2)$ is a repeated factor.

So, with least possible degree,

f(x)=a(x+2)^2(x-1)(x-4)

for some constant $a$ .

Use the labeled point to find $a$

The graph shows that $(0,2)$ lies on $f$ , so $f(0)=2$ .

Substitute $x=0$ :

2=a(0+2)^2(0-1)(0-4)=a\cdot 4\cdot (-1)\cdot (-4)=16a

So $a=\frac{1}{8}$ .

Evaluate $f(3)$

Now substitute $x=3$ :

\begin{aligned} f(3) &= \frac{1}{8}(3+2)^2(3-1)(3-4) \\ &= \frac{1}{8}(5^2)(2)(-1) \\ &= \frac{-50}{8} \\ &= -\frac{25}{4} \end{aligned}

Therefore, the value of $f(3)$ is $-\frac{25}{4}$ .

Question 130·Hard·Math

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