Question

question 3
1 pts
use your calculator to find the solutions and the maximum or minimum point of the quadratic equation.
y = -x² - 8x - 15
○ (-4,1) maximum value; x = -5 and x = -3
○ (-4,1) minimum value; no solution
○ (-4,-1) minimum value; x = -5 and x = -3
○ (4,-1) maximum value; no solution

Explanation:

Step1: Analyze the quadratic function

The quadratic function is \( y = -x^2 - 8x - 15 \). The coefficient of \( x^2 \) is -1, which is negative, so the parabola opens downward, meaning it has a maximum point.

Step2: Find the vertex (maximum point)

The x-coordinate of the vertex of a quadratic function \( y = ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \). Here, \( a = -1 \) and \( b = -8 \), so \( x = -\frac{-8}{2\times(-1)} = -4 \).
Substitute \( x = -4 \) into the function: \( y = -(-4)^2 - 8\times(-4) - 15 = -16 + 32 - 15 = 1 \). So the vertex (maximum point) is (-4, 1).

Step3: Find the solutions (x-intercepts)

Set \( y = 0 \), so \( -x^2 - 8x - 15 = 0 \). Multiply both sides by -1: \( x^2 + 8x + 15 = 0 \). Factor: \( (x + 5)(x + 3) = 0 \). So the solutions are \( x = -5 \) and \( x = -3 \).

Answer:

A. (-4,1) Maximum value; x = -5 and x = -3