How do you find the equation of the parabola that has its vertex at #(-2,5)# and #y#-intercept #9#?

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Answer 1 · 2016-09-11T06:46:54

#y=(x+2)^2+5#

Quadratic Equation in intercept form is -

#y=a(x-h)^2+k#

Where -

#h=-2#
#k=5#

The line is passing through the point #(0,9)#

Plug in the values of #h,k# in the formula

#y=a(x-(-2))^2+5#
#y=a(x+2)^2+5#

We have to find the value of #a#

The parabola is passing through the point #(0,9)#
Then

#a(0+2)^2+5=9#
#4a+5=9#
#4a=9-5#
#a=4/4=1#

plug in the value #a=1#in the equation #y=a(x+2)^2+5#

#y=(x+2)^2+5#

Look at the graph