Question

use the drawing tool(s) to form the correct answer on the provided graph.
plot the axis of symmetry and the point where the maximum value occurs for the function below.
$h(x) = -\frac{1}{2}x^2 - 2x + 6$

Explanation:

Step1: Find axis of symmetry

For a quadratic function \( h(x) = ax^2 + bx + c \), the axis of symmetry is \( x = -\frac{b}{2a} \). Here, \( a = -\frac{1}{2} \), \( b = -2 \). So \( x = -\frac{-2}{2\times(-\frac{1}{2})} = -\frac{-2}{-1} = -2 \).

Step2: Find maximum point

The maximum (since \( a < 0 \)) occurs at the vertex, which is on the axis of symmetry. Substitute \( x = -2 \) into \( h(x) \): \( h(-2) = -\frac{1}{2}(-2)^2 - 2(-2) + 6 = -\frac{1}{2}(4) + 4 + 6 = -2 + 4 + 6 = 8 \). So the point is \( (-2, 8) \).

Answer:

Axis of symmetry: \( x = -2 \) (a vertical line through \( x = -2 \) on the graph), Maximum point: \( (-2, 8) \) (plot this point on the graph).