Question

How do you find the vertex and the intercepts for `y = x^2 + 10x - 7`?

Answer 1

vertex (-5, -32)

Explanation:

y = x^2 + 10x - 7
x-coordinate of vertex:
x = -b/(2a) = -10/2 = -5
y-coordinate of vertex:
y(-5) = 25 - 50 - 7 = -32
To find x-intercepts, make y = 0 and solve the quadratic equation:
`x^2 + 10x - 7 = 0`
`D = d^2 = b^2 - 4ac = 100 + 28 = 128` --> `d = +- 8sqrt2`
There are 2real roots:
`x = -b/(2a) +- d/(2a) = -10/2 +- (8sqrt2)/2 = -5 +- 4sqrt2`
`x1 = -5 + 4sqrt2`
`x2 = -5 - 4sqrt2`