Question

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

Answer 1

`vertex (-1/2, -9/4)`

Explanation:

x-coordinate of vertex:
`x = -b/(2a) = - 1/2`
y-coordinate of vertex:
`y(-1/2) = 1/4 - 1/2 - 2 = - 9/4`
`Vertex (-1/2, -9/4)`.
To find y-intercept, make x = 0 --> y = -2.
To find x-intercepts, make y = 0 and solve the quadratic equation
`y = x^2 + x - 2 = 0`
Since a + b + c = 0, use shortcut.
The 2 real roots are 1 and `c/a = -2.`