# How do you find the vertex and intercepts for f(x)=12.25x^2 - 52.5x +110.25?

May 20, 2017

Vertex (2.14, 54)

#### Explanation:

$f \left(x\right) = 12.25 {x}^{2} - 52.5 x + 110.25$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{52.5}{24.5} = 2.14$
y-coordinate of vertex:
$f \left(2.14\right) = 12.25 {\left(2.14\right)}^{2} - 52.5 \left(2.14\right) + 110.25 =$
$= 56.1 - 112.35 + 110.25 = 54$
Vertex (2.14, 54)
To find the 2 intercepts, solve the quadratic equation:
$f \left(x\right) = 12.25 {x}^{2} - 52.5 x + 110.25 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 2756.25 - 5402.25 = - 2646 < 0$
Since D < 0, there are no x-intercepts. The parabola graph of f(x)
stays completely above the x-axis.