# How do you find the vertex and intercepts for y=5x^2-30x+49?

Apr 13, 2016

Vertex (3, 4)

#### Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{30}{10} = 3$
y-coordinate of vertex:
y(3) = 5(9) - 3(30) + 49 = 45 - 90 + 49 = 4
Vertex (3, 4).
Make x = 0, y-intercept = 49.
To find x-intercepts, make y = 0 and solve the quadratic equation:
$y = 5 {x}^{2} - 30 x + 49 = 0.$
$D = {b}^{2} - 4 a c = 900 - 980 = - 80.$
Since D < 0, there are no real roots (no x-intercepts). Since a > 0, the parabola opens upward and is completely above the x-axis.