# How do you find the vertex and intercepts for f(x) = 10x^2 - 5x + 12?

Jun 28, 2018

Vertex (1/4, 91/8)
No x-intercepts

#### Explanation:

$f \left(x\right) = 10 {x}^{2} - 5 x + 12$
x - coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{5}{20} = \frac{1}{4}$
y-coordinate of vertex:
$f \left(\frac{1}{4}\right) = 10 \left(\frac{1}{16}\right) - 5 \left(\frac{1}{4}\right) + 12 = \frac{5}{8} - \frac{10}{8} + \frac{96}{8} = \frac{91}{8}$
Vertex $\left(\frac{1}{4} , \frac{91}{8}\right)$
y-intercept when x = 0, --> f(0) = 12
x- intercepts when f(x) = 0. Solve the quadratic equation:
$f \left(x\right) = 10 {x}^{2} - 5 x + 12 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 25 - 480 = - 455 < 0.$
There are no x-intercepts. The graph of f(x) is an upward parabola (a > 0), that completely stays above the x-axis.