Question

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

1. $y = -2x^2 - 16x + 20$
2. $y = -x^2 + 12x - 15$

Explanation:

Problem 7: \( y = -2x^2 - 16x + 20 \)

Step 1: Identify the coefficient of \( x^2 \)

The coefficient of \( x^2 \) is \( a = -2 \). Since \( a < 0 \), the parabola opens downward, so it has a maximum value at the vertex.

Step 2: Find the x-coordinate of the vertex

The formula for the x-coordinate of the vertex of a parabola \( y = ax^2 + bx + c \) is \( x = -\frac{b}{2a} \). Here, \( a = -2 \) and \( b = -16 \).
\[
x = -\frac{-16}{2 \times (-2)} = -\frac{16}{4} = -4
\]

Step 3: Find the y-coordinate (maximum value)

Substitute \( x = -4 \) into the equation \( y = -2x^2 - 16x + 20 \):
\[
y = -2(-4)^2 - 16(-4) + 20 = -2(16) + 64 + 20 = -32 + 64 + 20 = 52
\]

Problem 8: \( y = -x^2 + 12x - 15 \)

Step 1: Identify the coefficient of \( x^2 \)

The coefficient of \( x^2 \) is \( a = -1 \). Since \( a < 0 \), the parabola opens downward, so it has a maximum value at the vertex.

Step 2: Find the x-coordinate of the vertex

Using the formula \( x = -\frac{b}{2a} \), where \( a = -1 \) and \( b = 12 \):
\[
x = -\frac{12}{2 \times (-1)} = -\frac{12}{-2} = 6
\]

Step 3: Find the y-coordinate (maximum value)

Substitute \( x = 6 \) into the equation \( y = -x^2 + 12x - 15 \):
\[
y = -(6)^2 + 12(6) - 15 = -36 + 72 - 15 = 21
\]

Answer:

• For problem 7: The maximum value of the parabola \( y = -2x^2 - 16x + 20 \) is \( \boldsymbol{52} \).
• For problem 8: The maximum value of the parabola \( y = -x^2 + 12x - 15 \) is \( \boldsymbol{21} \).