How do you find the quadratic function with vertex (-2,5) and point (0,9)?

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Answer 1 · 2018-01-07T23:51:12

#y=x^2+4x+9#

Vertex form of a quadratic equation is #y=a(x-h)^2+k# where #(h,k)# is the vertex

In this case, h is -2 and k is 5.

Thus the equation becomes #y=a(x+2)^2+5#

In order to find out a, we must plug in the x- and y-values of the point we want the equation to pass through.

The equation becomes:
#9=a(0+2)^2+5#

#=>9=a*2^2+5#

#=>4a=4#

#=>a=1#

Finally the equation is #y=(x+2)^2+5#

Another way to put it is:
#y=x^2+4x+9#