Question

How do you find the vertex and the intercepts for `y=x^(2)-2x-15`?

Answer 1

Vertex (1, -16)

Explanation:

x-coordinate of vertex:
`x = -b/(2a) = 2/2 = 1`
y-coordinate of vertex:
y(1) = 1 - 2 - 15 = - 16
Vertex (1, -16)
To find y-intercept, make x = 0 --> y = -15
To find x-intercept, make y = 0 and solve the quadratic equation:
`x^2 - 2x - 15 = 0.`
Find 2 real roots (x-intercepts) knowing sum (2) and product (-15).
They are x = -3 and x = 5