Question

How do you find the vertex and the intercepts for `y = 4x^2 -8x + 10`?

Answer 1

Vertex (1, 6)
No x-intercepts.

Explanation:

`y = 4x^2 - 8x + 10`
x-coordinate of vertex:
`x = -b/(2a) = 8/8 = 1`
y-coordinate of vertex:
y(1) = 4 - 8 + 10 = 6
Vertex( 1, 6).
To find the 2 x-intercepts, solve y = 0 by the quadratic formula.
`D = b^2 - 4ac = 64 - 160 = - 96`.
Because D < 0, there are no real roots (no x-intercepts). The parabola graph doesn't intersect with the x-axis. The parabola opens upward (a > 0). The parabola graph stays completely above the x-axis.
graph{4x^2 - 8x + 10 [-20, 20, -10, 10]}