Question

How do you find the vertex and intercepts for `y= -2x^2 + 12x - 13`?

Answer 1

Vertex (3, 5)

Explanation:

`y = - 2x^2 + 12x - 13`
x-coordinate of vertex:
`x = - b/(2a) = -12/-4 = 3`
y-coordinate of vertex:
y(3) = - 2(9) + 12(3) - 13 = - 18 + 36 - 13 = 5
Vertex (3, 5)
To find y-intercept, make x = 0 --> y-intercept = - 13.
To find the 2 intercept, make y = 0 and solve:
`y = - 2x^2 + 12x - 13 = 0`
`D = b^2 - 4ac = 144 - 104 = 40` --> `d = +- 2sqrt10`
There are 2 x-intercepts (real roots):
`x = - b/(2a) +- d/(2a) = -12/-4 +- (2sqrt10)/4 = 3 +- sqrt10/(2)`
graph{-2 x^2 + 12x - 13 [-20, 20, -10, 10]}