Question

graph the equation $y = -x^2 - 12x - 32$ on the accompanying set of axes. you must plot 5 points including the roots and the vertex. click to plot points. click points to delete them.

Explanation:

Step1: Find the roots (x-intercepts)

Set \( y = 0 \), so \( -x^2 - 12x - 32 = 0 \). Multiply both sides by -1: \( x^2 + 12x + 32 = 0 \). Factor: \( (x + 4)(x + 8) = 0 \). So roots are \( x = -4 \) and \( x = -8 \). Thus, points \((-4, 0)\) and \((-8, 0)\).

Step2: Find the vertex

For quadratic \( y = ax^2 + bx + c \), vertex x-coordinate is \( x = -\frac{b}{2a} \). Here, \( a = -1 \), \( b = -12 \). So \( x = -\frac{-12}{2(-1)} = -6 \). Plug \( x = -6 \) into equation: \( y = -(-6)^2 - 12(-6) - 32 = -36 + 72 - 32 = 4 \). Vertex is \((-6, 4)\).

Step3: Find two more points

Choose \( x = -5 \): \( y = -(-5)^2 - 12(-5) - 32 = -25 + 60 - 32 = 3 \). Point \((-5, 3)\).

Choose \( x = -7 \): \( y = -(-7)^2 - 12(-7) - 32 = -49 + 84 - 32 = 3 \). Point \((-7, 3)\).

Answer:

Points to plot: \((-8, 0)\), \((-6, 4)\), \((-4, 0)\), \((-5, 3)\), \((-7, 3)\) (plot these on the axes).