Question

How do you find the vertex and intercepts for `y = 8x + 5x^2 – 3`?

Answer 1

Vertex `(-4/5, -31/5)`

Explanation:

`y = 5x^2 + 8x - 3`
x-coordinate of vertex:
`x = -b/(2a) = -8/10 = -4/5`
y-coordinate of vertex:
`y(-4/5) = 5(16/25) - 32/5 - 3 = 16/5 - 32/5 - 3 = - 31/5`
Vertex `(-4/5, -31/5)`
To find y-intercept, make x = 0 --> y = -3
To find 2 x-intercepts, make y = 0 and solve the quadratic equation `y = 5x^2 + 8x - 3 = 0`
`D = d^2 = b^2 - 4ac = 64 + 60 = 124` --> `d = +- 2sqrt31`
There are 2 x-intercepts (2 real roots):
`x = -b/(2a) +- d/(2a) = -8/10 +- (2sqrt31)/10 = -4/5 +- sqrt31/5`
`x = (-4 +- sqrt31)/5`