Question

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

Answer 1

The graph is a parabola that opens downward.

Generally, a parabola has the following equation:

`y = ax^2 + bx + c` If a is positive it opens upward. If a is negative it opens downward.

`y= -4x^2 +20x -24`
`a= -4, b = 20, c = -24`

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

`h= 5/2`

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

`y = -4(5/2)^2 +20(5/2) -24`

`y = 1` which is `k`

The vertex is at `(5/2,1)`

II. x and y intercepts

set y = 0 to get the x-intercepts

`0 = -4x^2 + 20x - 24`
divide both sides by -4

`0 = x^2 -5x + 6`, then factor

`0 = (x-3)(x-2)`

`x= 3 , x =2`

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

`y = -4(0)^2 +20(0) - 24`

`y= -24`

How to graph?
1. Plot the vertex at `( 5/2,1)`. This is where opening downward commences.
2. Plot the x intercepts `(3,0), (2,0)` this is where the graph will cross the x-axis.
3. Plot the y-intercept `(0,-24)` 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.