find the x-intercept of the parabola with vertex (4 -1) and y-intercept (0 15) write your answer in this form (x1,y1) , (x2,y2)

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

y=a(x-4)^2-1

The y intercept given allows you this relation:

15=a(0-4)^2-1

15=16a-1

16=16a

a=1

then y=(x-4)^2-1

x-intercept of the parabola:

(x-4)^2-1=0

(x-4)^2=1

x-4=-1 or x-4=1

x=3 or x=5

so x-intercept are (3, 0) and (5, 0)

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