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

Apr 29, 2016

Vertex (2, 5)

#### Explanation:

$f \left(x\right) = 4 {x}^{2} - 16 x + 21$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{16}{8} = 2.$
y-coordinate of vertex:
$y \left(2\right) = 16 - 32 + 21 = 5$
Vertex (2, 5)
To find y-intercept, make x = 0 --> y = 21
To find x-intercepts, make y = 0 and solve the quadratic equation:
$y = 4 {x}^{2} - 16 x + 21 = 0$
$D = {b}^{2} - 4 a c = 256 - 336 = - 80.$
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.