Find the equation of parabola whose axis of symmetry is #x=-3# and passes through points #(-1,3)# and #(2,-5)#?

1 Answer

Answer:

Please see below.

Explanation:

Given that the axis is #x = -3#, we know the vertex is at #x = -3#, so the equation can be written in the form

#y = a(x+3)^2 +k#

Now use the two points given to get:

From #(-1,3)#, we have #3 = a(2)^2+k# so #4a+k = 3# and #k = 3-4a#

From #(2,-5)#, we get #-5 = a(5)^2+k# so #25a+k=-5#

Replace #k# in the last equation with #3-4a# to get

#25a+(3-4a) = -5#, so

#21a = -8# and # a = -8/21#

Using #k = 3-4a# again we get #k = 3-(-32/21) = 63/21+32/21 = 95/21#.

The equation is #y = -8/21(x+3)^2 + 95/21#

If course, there are other ways to write the answer.

graph{(y+8/21(x+3)^2-95/21)(x+3)((x+1)^2+(y-3)^2-0.02)((x-2)^2+(y+5)^2-0.02)=0 [-12, 12, -6, 6]}