Question

Find the x-intercepts of the parabola with the vertex (4,-1) and y-intercpet (0,15). write your answer in this form, (x1,y1)(x2.y2)
use this method (y-k)=a(x-h)^2 to solve the problem

Answer 1

If you fit the vertex and the y-intercept into that form of the parabola, what you get is this:
$a(0-4) ^{2} =(15+1)$ and
$a(-4) ^{2} =16$ and 16a=16. Solve that a to get that a = 1.  Now you have your true parabola equation in vertex form, which is:
$1(x-4) ^{2} =y+1$ or just
$(x-4) ^{2} =y+1$
Follow my work here to find the x-intercepts:
$(x-4) ^{2} -1=y$
(x-4)(x-4)-1=y and $x^{2} -8x+16-1=y$
Simplifying by combining like terms gives you this parabola:
$x^{2} -8x+15=y$
In order to find the x-intercepts, or the roots of the parabola as they are also known as, you have to set y equal to 0 (because the x-intercepts are found when y = 0, right?)
$x^{2} -8x+15=0$
Factoring that gives you (x-5)(x-3)=0 and that x-5=0 and x-3=0. Solving each of those simple equations gives you x=5 and x=3 when y=0. So the roots, in the form you require them, are (5, 0) and (3, 0)