Question

How do you find the vertex and the intercepts for `y = -2x^2 + 13x - 6`?

Answer 1

`Vertex(13/4, 121/8)`

Explanation:

x-coordinate of vertex:
`x = -b/(2a) = -13/-4 = 13/4`
y-coordinate of vertex:
`y(13/4) = -2(169/16) + 13(13/4) - 6 = 169/8 - 6 = 121/8`
Vertex `(13/4, 121/8)`
To find y-intercept, make x = 0 --> y = -6.
To find x-intercepts, make y = 0 and solve the quadratic equation
`-2x^2 + 13x - 6 = 0`
`D = d^2 = b^2 - 4ac = 169 - 48 = 121` --> `d = +- 11`
There are 2 real roots (x-intercepts):
`x = - b/(2a) +- d/(2a) = 13/4 +- 11/4`
graph{-2x^2 + 13x - 6 [-40, 40, -20, 20]}