Question

How do you find the vertex and the intercepts for `y=x^2 - 8x + 3`?

Answer 1

Vertex (4, -13)

Explanation:

`y = x^2 - 8x + 3`
x-coordinate of vertex:
`x = -b/(2a) = 8/2 = 4`
y-coordinate of vertex:
y(4) = 16 - 32 + 3 = -13
Vertex (4, -13)
To find y-intercepts, make x = 0 --> y = 3
To find x-intercepts, solve the quadratic equation y = 0
Use the improved quadratic formula (Google Search)
`D = d^2 = b^2 - 4ac = 64 - 12 = 52` --> `d = +- 2sqrt13`
There are 2 x-intercepts (2 real roots):
`x = -b/(2a) +- d/(2a) = 8/2 +- (2sqrt13)/2 = 4 +- sqrt13`
graph{x^2 - 8x + 3 [-40, 40, -20, 20]}