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

Nov 24, 2015

Find vertex and intercepts of y = x^2 - 4x - 12

Explanation:

x-coordinate of vertex: $x = \left(- \frac{b}{2 a}\right) = \frac{4}{2} = 2$
y-coordinate of vertex: y = y(2) = 4 - 8 - 12 = -16
To find y-intercept, make x = 0 --> y = -12
To find x-intercepts, make y = 0 and solve the quadratic equation:
$y = {x}^{2} - 4 x - 12 = 0$.
Find 2 numbers knowing sum (4) and product (-12).
Factor pairs of (-12) --> (-2, 6). This sum is 4 = -b. Then the 2 x-intercepts (real roots) are: -2 and 6.