Question

the function $f(x) = -x^2 + 6x + 16$ represents the net profit in thousands of dollars, $f(x)$, for a new product $x$ months after its launch.

what is the equivalent form of the function that shows the months when the profit is equal to zero?

a. $f(x) = -(x - 3)^2 + 25$

b. $f(x) = (x - 3)^2 + 25$

c. $f(x) = (x + 2)(x - 8)$

d. $f(x) = -(x + 2)(x - 8)$

Explanation:

Step1: Understand the goal

We need to find the equivalent form of \( f(x) = -x^2 + 6x + 16 \) that shows the months when profit is zero (i.e., factor the quadratic or find its factored form, since profit zero means \( f(x) = 0 \), so we need to solve \( -x^2 + 6x + 16 = 0 \) or factor the quadratic).

Step2: Factor the quadratic (or manipulate it)

First, let's factor out the negative sign from the first two terms:
\( f(x) = - (x^2 - 6x) + 16 \)

To complete the square inside the parentheses (though we can also factor the quadratic directly). Let's try factoring \( -x^2 + 6x + 16 \). Multiply both sides by -1: \( x^2 - 6x - 16 \). Now, we need two numbers that multiply to -16 and add to -6. Wait, no: original quadratic is \( -x^2 + 6x + 16 \), so let's factor it as \( - (x^2 - 6x - 16) \). Now factor \( x^2 - 6x - 16 \): find two numbers that multiply to -16 and add to -6? Wait, no, wait: \( x^2 - 6x - 16 \). Wait, actually, let's use the quadratic formula or factor. Let's try factoring \( x^2 - 6x - 16 \). We need two numbers \( a \) and \( b \) such that \( a \times b = -16 \) and \( a + b = -6 \). Wait, no, wait: \( x^2 - 6x - 16 \). Wait, maybe I made a mistake. Wait, original function: \( -x^2 + 6x + 16 \). Let's set \( f(x) = 0 \): \( -x^2 + 6x + 16 = 0 \) → multiply both sides by -1: \( x^2 - 6x - 16 = 0 \). Now, factor \( x^2 - 6x - 16 \). Let's find two numbers that multiply to -16 and add to -6. Wait, 2 and -8: 2 * (-8) = -16, 2 + (-8) = -6. So, \( x^2 - 6x - 16 = (x + 2)(x - 8) \). But wait, our original function is \( - (x^2 - 6x - 16) \), so substituting back: \( f(x) = - (x + 2)(x - 8) \). Let's check:

Expand \( - (x + 2)(x - 8) \):

First, multiply \( (x + 2)(x - 8) = x^2 - 8x + 2x - 16 = x^2 - 6x - 16 \). Then multiply by -1: \( -x^2 + 6x + 16 \), which matches the original function. So the factored form is \( f(x) = - (x + 2)(x - 8) \), which is option D.

Alternatively, let's check the options:

Option A: \( - (x - 3)^2 + 25 \). Let's expand this: \( - (x^2 - 6x + 9) + 25 = -x^2 + 6x - 9 + 25 = -x^2 + 6x + 16 \). So this is the vertex form, which shows the maximum profit (vertex), but not the roots (when profit is zero). So A is vertex form, not factored form.

Option B: \( (x - 3)^2 + 25 \). Expand: \( x^2 - 6x + 9 + 25 = x^2 - 6x + 34 \), which is not equal to original function (original has -x², so B is wrong).

Option C: \( (x + 2)(x - 8) \). Expand: \( x^2 - 8x + 2x - 16 = x^2 - 6x - 16 \), which is not equal to original function (original is -x² +6x +16, so C is missing the negative sign).

Option D: \( - (x + 2)(x - 8) \). As we expanded earlier, this is equal to \( -x^2 + 6x + 16 \), which is the original function. And this factored form will allow us to set \( f(x) = 0 \) and solve \( - (x + 2)(x - 8) = 0 \), so \( x + 2 = 0 \) or \( x - 8 = 0 \), giving \( x = -2 \) (discarded, since months can't be negative) or \( x = 8 \). So this form shows the months when profit is zero.

Answer:

D. \( f(x) = - (x + 2)(x - 8) \)