Question

The function f(x) = x3 – 8x2 + x + 42 has zeros located at 7, –2, 3. Verify the zeros of f(x) and explain how you verified them. Describe the end behavior of the function.

Answer 1

• zeros are {-2, 3, 7} as verified by graphing
• end behavior: f(x) tends toward infinity with the same sign as x

Explanation:

A graphing calculator makes finding or verifying the zeros of a polynomial function as simple as typing the function into the input box.

Zeros

The attachment shows the function zeros to be x ∈ {-2, 3, 7}, as required.

End behavior

The leading coefficient of this odd-degree polynomial is positive, so the value of f(x) tends toward infinity of the same sign as x when the magnitude of x tends toward infinity.

• x → -∞; f(x) → -∞
• x → ∞; f(x) → ∞

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Additional comment

The function is entered in the graphing calculator input box in "Horner form," which is also a convenient form for hand-evaluation of the function.

We know the x^2 coefficient is the opposite of the sum of the zeros:

-(7 +(-2) +3) = -8 . . . . x^2 coefficient

And we know the constant is the opposite of the product of the zeros:

-(7)(-2)(3) = 42 . . . . . constant

These checks lend further confidence that the zeros are those given.

(The constant is the opposite of the product of zeros only for odd-degree polynomials. For even-degree polynomials. the constant is the product of zeros.)