Question

How do you identify the important parts of `y = –2x^2 – 32x – 126` to graph it?

Answer 1

Graph `y = -2x^2 - 32x - 126`

Explanation:

The important parts to find are:
x-coodinate of vertex and axis of symmetry:
`x = (-b/2a) = 32/4 = 8`
y-coordinate of vertex:
y = f(8) = -2(64) - 32(-8) - 126 = 2
y-intercept --> Make x = 0 --> y = -126
x-intercepts --> Make x = 0, solve y = 0
`y = -2(x^2 + 16x + 63) = 0.`
Roots have same sign, Factor pairs of (63) --> (3, 31)(7, 9). This sum is 16 = b. Then the 2 x-intercepts (real roots) are : 7 and 9
graph{-2x^2 - 32x - 126 [-10, 10, -5, 5]}