Question

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

Answer 1

"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]