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

Jul 9, 2017

$y = - {x}^{2} + 4 x + 12$
x-coordinate of vertex;
$x = - \frac{b}{2} a = - \frac{4}{-} 2 = 2$
y-coordinate of vertex:
$y \left(2\right) = - 4 + 8 + 12 = 16$
Vertex (2, 16)
Make x = 0 --> y-intercept = 12
Make y = 0, and solve the quadratic equation:
$y = - {x}^{2} + 4 x + 12 = 0$
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]}