Question

11 what is the vertex and axis of symmetry of the equation? $y = -2x^2 + 8x - 20$ a vertex: $(2, -12)$ axis of symmetry: $x = 2$ b vertex: $(2, -12)$ axis of symmetry: $x = -12$ c vertex: $(-2, -12)$ axis of symmetry: $x = -2$ d vertex: $(-2, 12)$ axis of symmetry: $y = -2$

Explanation:

Step1: Find x-coordinate of vertex

For quadratic \( y = ax^2 + bx + c \), vertex x - coord is \( x = -\frac{b}{2a} \). Here, \( a = -2 \), \( b = 8 \). So \( x = -\frac{8}{2\times(-2)} = 2 \).

Step2: Find y-coordinate of vertex

Substitute \( x = 2 \) into \( y = -2x^2 + 8x - 20 \). \( y = -2(2)^2 + 8(2) - 20 = -8 + 16 - 20 = -12 \). So vertex is \( (2, -12) \).

Step3: Axis of symmetry

Axis of symmetry for parabola \( y = ax^2 + bx + c \) is \( x = -\frac{b}{2a} \), which we found as \( x = 2 \).

Answer:

A. vertex: \( (2, -12) \)
axis of symmetry: \( x = 2 \)