Question

for $y = -x^2 - 8x - 7$
(a) find the vertex.
(b) find the $y$-intercept.
(c) find the $x$-intercept(s), if they exist.
(d) use this information to graph the function.

Explanation:

Part (a): Find the vertex

Step1: Recall vertex formula for quadratic

For \( y = ax^2 + bx + c \), vertex \( x \)-coordinate is \( x = -\frac{b}{2a} \). Here, \( a = -1 \), \( b = -8 \).
\( x = -\frac{-8}{2(-1)} = -\frac{8}{2} = -4 \)

Step2: Find \( y \)-coordinate of vertex

Substitute \( x = -4 \) into \( y = -x^2 - 8x - 7 \):
\( y = -(-4)^2 - 8(-4) - 7 = -16 + 32 - 7 = 9 \)

Step1: Set \( x = 0 \) in the equation

Substitute \( x = 0 \) into \( y = -x^2 - 8x - 7 \):
\( y = -(0)^2 - 8(0) - 7 = -7 \)

Step1: Set \( y = 0 \) and solve \( -x^2 - 8x - 7 = 0 \)

Multiply both sides by \(-1\): \( x^2 + 8x + 7 = 0 \)

Step2: Factor the quadratic

\( x^2 + 8x + 7 = (x + 1)(x + 7) = 0 \)

Step3: Solve for \( x \)

\( x + 1 = 0 \) or \( x + 7 = 0 \) gives \( x = -1 \) or \( x = -7 \)

Answer:

(a):
The vertex is \((-4, 9)\)

Part (b): Find the \( y \)-intercept