# How do you find the vertex and intercepts for f(x) = -5x^2 + 7x + 4?

Dec 2, 2015

Find vertex and intercepts of f(x) = - 5x^2 + 7x + 4

#### Explanation:

x-coordinate of vertex:
$x = \frac{- b}{2 a} = \frac{7}{10}$
y-coordinate of vertex:
$y = f \left(\frac{7}{10}\right) = - \frac{49}{100} + \frac{49}{10} + 4 = \frac{645}{100}$
To find y-intercept, make x = 0 --> f(x) = 4
To find x-intercepts, make y = 0 and solve the quadratic equation:
$- 5 {x}^{2} + 7 x + 4 = 0$.
$D = {d}^{2} = {b}^{2} - 4 a c = 49 + 80 = 129$ --> $d = \pm \sqrt{29} = \pm 5.39$
$x = - \frac{7}{10} \pm \frac{5.39}{10} = - 0.7 - 5.39 = - 6.09$ and
$x = - \frac{7}{10} + 5.39 = 5.39 - 0.7 = 4.69$