Question

find the y-intercept, the axis of symmetry, and the vertex of the graph of the function.
$f(x) = -x^2 - 2x - 2$

the y-intercept is: \boxed{} (type an ordered pair.)

the axis of symmetry is: \boxed{} (simplify your answer. type an equation.)

the vertex is: \boxed{} (type an ordered pair.)

Explanation:

Step1: Identify the function

The function is \( f(x) = -x^2 - 2x - 3 \). To find the y - intercept, we set \( x = 0 \).
Substitute \( x = 0 \) into the function: \( f(0)=-(0)^2-2(0)-3=-3 \). So the y - intercept is \( (0, - 3) \).

Step2: Rewrite the function in vertex form (for axis of symmetry and vertex)

First, complete the square for the quadratic function \( y=-x^{2}-2x - 3 \).
Factor out - 1 from the first two terms: \( y=-(x^{2}+2x)-3 \).
To complete the square inside the parentheses, we take half of the coefficient of \( x \) (which is 2), square it (\( (\frac{2}{2})^2 = 1 \)), and add and subtract it inside the parentheses:
\( y=-(x^{2}+2x + 1-1)-3 \)
\( y=-( (x + 1)^{2}-1)-3 \)
\( y=-(x + 1)^{2}+1 - 3 \)
\( y=-(x + 1)^{2}-2 \)

Step3: Find the axis of symmetry and vertex

For a quadratic function in the form \( y = a(x - h)^{2}+k \), the axis of symmetry is \( x=h \) and the vertex is \( (h,k) \).
From \( y=-(x + 1)^{2}-2 \), we can rewrite it as \( y=-(x-(-1))^{2}+(-2) \). So \( h=-1 \) and \( k = - 2 \).
The axis of symmetry is \( x=-1 \) and the vertex is \( (-1,-2) \).

Answer:

• The y - intercept is \( (0, - 3) \)
• The axis of symmetry is \( x=-1 \)
• The vertex is \( (-1, - 2) \)