Question

How do you find the vertex and the intercepts for `f(x) = -x^2 + x - 5`?

Answer 1

Vertex: `=(1/2, 5 3/4)`
`y`-intercept: `x=0`, `y=-5`
`x`-Intercept: none

Explanation:

This equation is in standard form
#y=ax^2+bx+c

Where `a=-1`, `b=1`, and `c=-5`

Vertex: `(-b/2a, f (-b/2a))`
`=(1/2, 5 3/4)`

Intercepts:
When `x=0`, `y=-5`
When `y=0`, you have to use the quadratic equation. But the number under the square root is negative, so there are no `x`-Intercepts.