What is the equation of the parabola that has a vertex at # (-4, 4) # and passes through point # (6,104) #?

1 Answer
Mar 4, 2016

#y=(x+4)^2 + 4# or
#y=x^2 + 8*x + 20#

Explanation:

Start with the vertex form of the quadratic equation.

#y=a*(x-x_{vertex})^2 + y_{vertex}#.

We have #(-4,4)# as our vertex, so right off the bat we have

#y=a*(x-(-4))^2 + 4# or

#y=a*(x+4)^2 + 4#, less formally.

Now we just need to find "#a#."
To do this we sub in the values for the second point #(6,104)# into the equation and solve for #a#.
Subbing in we find
#(104)=a*((6)+4)^2 + 4#
or
#104=a*(10)^2+4#.

Squaring #10# and subtracting #4# from both sides leaves us with
#100=a*100# or #a=1#.

Thus the formula is #y=(x+4)^2 + 4#.

If we want this in standard form ( #y=a*x^2 + b*x +c # ) we expand the square term to get

#y=(x^2 + 8*x + 16) + 4# or

#y=x^2 + 8*x + 20#.