Parabola Vertex (1,5) y intercept (0,2)
First we fix the vertex:
[tex]y = a(x - 1)^2 + 5[/tex]
The y intercept lets us solve for a
[tex]2 = a(0 - 1)^2 + 5 = a+5[/tex]
[tex]a=-3[/tex]
So our parabola is
[tex]y = -3(x - 1)^2 + 5[/tex]
The x intercepts are the zeros of the polynomial. We've already completed the square, so let's solve directly for the zeros:
[tex]0 = -3(x - 1)^2 + 5[/tex]
[tex]3(x - 1)^2 = 5[/tex]
[tex](x - 1)^2 = \frac 5 3[/tex]
[tex]x - 1 = \pm \sqrt{\frac5 3}[/tex]
[tex]x = 1 \pm \sqrt{\frac53}[/tex]
So our x intercepts are
[tex](1- \sqrt{\frac53}, 0) \quad \textrm{and} \quad (1+\sqrt{\frac53}, 0)[/tex]
I hate ruining a nice exact answer with an approximation so I'll leave the calculator part to you.
