# How do you find the vertex, x-intercept, y-intercept, and graph the equation y=-4x^2+20x-24?

Dec 17, 2014

The graph is a parabola that opens downward.

Generally, a parabola has the following equation:

$y = a {x}^{2} + b x + c$ If a is positive it opens upward. If a is negative it opens downward.

$y = - 4 {x}^{2} + 20 x - 24$
$a = - 4 , b = 20 , c = - 24$

I. Solve for Vertex $\left(h , k\right)$ Use the formula  h= -b/(2a) =-20/(2(-4)

$h = \frac{5}{2}$

To solve for k plug in the value of h to x in the original equation

$y = - 4 {\left(\frac{5}{2}\right)}^{2} + 20 \left(\frac{5}{2}\right) - 24$

$y = 1$ which is $k$

The vertex is at $\left(\frac{5}{2} , 1\right)$

II. x and y intercepts

set y = 0 to get the x-intercepts

$0 = - 4 {x}^{2} + 20 x - 24$
divide both sides by -4

$0 = {x}^{2} - 5 x + 6$, then factor

$0 = \left(x - 3\right) \left(x - 2\right)$

$x = 3 , x = 2$

To get the y-intercepts, set $x = 0$

$y = - 4 {\left(0\right)}^{2} + 20 \left(0\right) - 24$

$y = - 24$

How to graph?
1. Plot the vertex at $\left(\frac{5}{2} , 1\right)$. This is where opening downward commences.
2. Plot the x intercepts $\left(3 , 0\right) , \left(2 , 0\right)$ this is where the graph will cross the x-axis.
3. Plot the y-intercept $\left(0 , - 24\right)$ This is where the graph crosses the y-axis.
Can anyone help me to create a picture of the graph here. I really don't know how. Thanks.