In the xy-plane, the graph of the polynomial function f has x-intercepts at (-3, 0), (1, 0… — SAT

In the $xy$ -plane, the graph of the polynomial function $f$ has $x$ -intercepts at $(-3, 0)$ , $(1, 0)$ , and $(k, 0)$ , where $k$ is a constant greater than $1$ . The function is defined by $f(x) = -2(x + 3)(x - 1)(x - k)$ . If the graph of $f$ passes through the point $(2, 20)$ , what is the value of $k$ ?

Answer: 4

Answer

The value of

k

4

Substituting the coordinates of the point

(2, 20)

into the function equation

f(x) = -2(x + 3)(x - 1)(x - k)

gives

20 = -2(2 + 3)(2 - 1)(2 - k)

. Simplifying this expression yields

20 = -10(2 - k)

. Dividing both sides by

-10

gives

-2 = 2 - k

, which simplifies to

k = 4

. Since

4 > 1

, this meets the condition that

k

is a constant greater than

1

Step-by-Step Solution

Substitute the coordinates of the point

(2, 20)

into the function definition.

20 = -2(2 + 3)(2 - 1)(2 - k)

Since the graph passes through the point

(2, 20)

, the coordinates satisfy the equation of the function.

Simplify the numerical factors on the right side of the equation.

20 = -10(2 - k)

Calculating

2+3 = 5

and

2-1 = 1

, then multiplying the constants:

-2 \times 5 \times 1 = -10

Solve the linear equation for

k

k = 4

Dividing both sides by

-10

yields

2 - k = -2

. Adding

k

to both sides and adding

2

to both sides gives

k = 4

Key Concept

Determining a constant root of a polynomial function by evaluating it at a given point on its graph.

Estimated Time:1m 30s

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