Question

How do you find the vertex and intercepts for `y=-x^2+4x+12`?

Answer 1

`y = - x^2 + 4x + 12`
x-coordinate of vertex;
`x = - b/2a = -4/-2 = 2`
y-coordinate of vertex:
`y(2) = - 4 + 8 + 12 = 16`
Vertex (2, 16)
Make x = 0 --> y-intercept = 12
Make y = 0, and solve the quadratic equation:
`y = - x^2 + 4x + 12 = 0`
Find 2 real roots knowing the sum (b = 4) and product (ac = - 12). The 2 x-intercepts are: x = - 2 and x = 6.
graph{- x^2 + 4x + 12 [-40, 40, -20, 20]}