Question

How do you find the vertex and intercepts for `y = x^2-8x+7`?

Answer 1

vertex (4, -9)

Explanation:

x-coordinate of vertex:
`x = -b/(2a) = 8/2 = 4`
y-coordinate of vertex"
`y(4) = 16 - 32 + 7 = -9`
y-intercept --> x = 0 --> y = 7
x-intercepts --> y = 0 --> solve `x^2 - 8x + 7 = 0.`
Since a + b + c = 0, use shortcut. --> x = 1 and `x = c/a = 7`
graph{x^2 - 8x + 7 [-20, 20, -10, 10]}