Question

for 7 and 8, find the maximum or minimum of the parabola.

1. $y = -2x^2 - 16x + 20$

Explanation:

Step1: Determine the vertex x - coordinate

For a quadratic function \( y = ax^{2}+bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). In the function \( y=-2x^{2}-16x + 20 \), \( a=-2 \) and \( b = - 16 \).
\[x=-\frac{-16}{2\times(-2)}=-\frac{-16}{-4}=-4\]

Step2: Find the y - coordinate of the vertex

Substitute \( x = - 4 \) into the function \( y=-2x^{2}-16x + 20 \).
\[

$$\begin{align*} y&=-2\times(-4)^{2}-16\times(-4)+20\\ &=-2\times16 + 64+20\\ &=-32 + 64+20\\ &=32 + 20\\ &=52 \end{align*}$$

\]

Step3: Determine if it's a maximum or minimum

Since \( a=-2<0 \), the parabola opens downwards, so the vertex is a maximum point.

Answer:

The maximum value of the parabola \( y = - 2x^{2}-16x + 20 \) is \( 52 \) (occurring at \( x=-4 \)).