Question

How do you find the vertex and the intercepts for `f(x) = 4x^2 - 16x +21`?

Answer 1

Vertex (2, 5)

Explanation:

`f(x) = 4x^2 - 16x + 21`
x-coordinate of vertex:
`x = -b/(2a) = 16/8 = 2.`
y-coordinate of vertex:
`y(2) = 16 - 32 + 21 = 5`
Vertex (2, 5)
To find y-intercept, make x = 0 --> y = 21
To find x-intercepts, make y = 0 and solve the quadratic equation:
`y = 4x^2 - 16x + 21 = 0`
`D = b^2 - 4ac = 256 - 336 = - 80.`
Since D < 0, there are no real roots, no x-intercepts.
Since a > 0 the parabola opens upward and it is completely above the x-axis.