Question

How do find the vertex and axis of symmetry, and intercepts for a quadratic equation `y=x^2-7x-28`?

Answer 1

Find vertex of y = x^2 - 7x - 28

Explanation:

x of vertex: `x = (-b/2a) = 7/2`

y of vertex: `y = f(7/2) = 49/4 - 49/2 - 28 = 161/4`

x of axis of symmetry = x of vertex =`7/2`

To find y-intercept, make x = 0 -> y = -28
To find x-intercepts, make y = 0
`D = d^2 = b^2 - 4ac` = 49 + 112 = 161 -->`d = +- sqrt161`

x-intercepts:`x = 7/2 +- (sqrt161)/2`