Question

1. sketch a parabola with an axis of symmetry x = -1, y-intercept 1, and what is the maximum or minimum value of the function? what is the range? 14. y = 2x² + 32x - 2 what is the vertex form of the equation?

Explanation:

Step1: Recall Vertex Form Conversion

To convert \( y = 2x^2 + 32x - 2 \) to vertex form \( y = a(x - h)^2 + k \), we complete the square. First, factor out the coefficient of \( x^2 \) from the first two terms: \( y = 2(x^2 + 16x) - 2 \).

Step2: Complete the Square

Take half of 16 (which is 8), square it (\( 8^2 = 64 \)), and add/subtract it inside the parentheses. But since we factored out a 2, we actually add \( 2 \times 64 \) and subtract \( 2 \times 64 \) to keep the equation balanced: \( y = 2(x^2 + 16x + 64 - 64) - 2 \).

Step3: Rewrite as Perfect Square

Rewrite \( x^2 + 16x + 64 \) as \( (x + 8)^2 \): \( y = 2((x + 8)^2 - 64) - 2 \).

Step4: Distribute and Simplify

Distribute the 2: \( y = 2(x + 8)^2 - 128 - 2 \), then simplify: \( y = 2(x + 8)^2 - 130 \).

Answer:

The vertex form of the equation \( y = 2x^2 + 32x - 2 \) is \( y = 2(x + 8)^2 - 130 \).