How do you find the x-intercepts of the parabola with vertex (4,-1) and y-intercept of (0,15)?

1 Answer
Dec 21, 2016

"x"-intercepts of the parabola are #(3,0) and (5,0)#

Explanation:

The equation of parabola with vertex # (h,k) or (4, -1)# is #y = a(x-h)^2+k or y = a(x-4)^2 -1 #. The parabola passes through point #(0,15)#. So the point will satisfy the equation of parabola.
#15=a(0-4)^2-1 or 15=16a-1 or 16a=16 or a=1#
Hence the equation of parabola is #y = (x-4)^2 -1 #. To find x -intercepts putting #y=0# in the equation we get #0 = (x-4)^2 -1 or x^2-8x+15=0 or (x-3)(x-5)=0 :. x=3 or x=5#. So x-intercepts are #(3,0) and (5,0)# graph{(x-4)^2-1 [-11.25, 11.25, -5.625, 5.625]}[Ans]