How do you write an equation in standard form of the parabola that has vertex (-8,-3) and passes through the point (4,717)?

1 Answer
Feb 20, 2018

Answer:

#y=5x^2+80x+317#

Explanation:

The formula for a parabola can be written as #y=a(x-h)^2+k#, where #(h,k)# is the vertex of the parabola.

Here, #h=-8# and #k=-3#.

We can input the above into the equation:

#y=a(x-(-8))^2+(-3)#, which simplifies to:

#y=a(x+8)^2-3#

We know that one of the coordinates is #(4,717)#. So when #x=4, y=717#. We need to find #a#, and so we can input:

#717=a(4+8)^2-3#

#717=a(12^2)-3#

#717=144a-3#

#144a=720#

#a=720/144#

#a=5#

Since #a=5#, we can input this into #y=a(x+8)^2-3#.

#y=5(x+8)^2-3#

#y=5(x^2+16x+64)-3#

#y=5x^2+80x+320-3#

#y=5x^2+80x+317#, the formula for the parabola.

We can graph it:

graph{5x^2+80x+317 [-29.24, 35.7, -11.13, 21.33]}

Notice the vertex is at #(-8,-3)#.

So the above is proved.