Question

a quadratic function f(x) is hidden from view. you must find the y-intercept(s) of f(x) and write the answer(s) in the form (x, y). choose the form of the quadratic function f(x) that you would like to see in order to answer the question most efficiently. form: vertex form f(x) = -(x + 1)^2 + 16 answer attempt 1 out of 2 there are no y-intercepts there are no y-intercepts there is one y-intercept there are two y-intercepts

Explanation:

Step1: Recall y-intercept definition

The y - intercept of a function \( f(x) \) is the point where \( x = 0 \). So we substitute \( x=0 \) into the function \( f(x)=-(x + 1)^{2}+16 \).

Step2: Substitute \( x = 0 \) into the function

Substitute \( x = 0 \) into \( f(x)=-(x + 1)^{2}+16 \):

$$
LATEXBLOCK0
$$

So when \( x = 0 \), \( y=15 \), which means there is one y - intercept at the point \( (0,15) \). Wait, but the original dropdown has a wrong option selected. Let's correct it. The function is a quadratic, and for any quadratic function \( y = ax^{2}+bx + c \), there is exactly one y - intercept (since substituting \( x = 0 \) gives \( y=c \), and a quadratic is a polynomial, defined for all real \( x \), so when \( x = 0 \), there is one value of \( y \)). So first, we find the y - intercept by plugging \( x = 0 \) into the vertex form.

Wait, let's re - do the substitution:

Given \( f(x)=-(x + 1)^{2}+16 \), expand it first maybe? \( f(x)=-(x^{2}+2x + 1)+16=-x^{2}-2x - 1 + 16=-x^{2}-2x+15 \). Now, the y - intercept is when \( x = 0 \), so \( f(0)=15 \), so the y - intercept is \( (0,15) \), so there is one y - intercept.

Answer:

There is one y - intercept \((0, 15)\)