Question

Find the x-intercepts intercept of the parabola

Answer 1

Parabola Vertex (1,5) y intercept (0,2)

First we fix the vertex:


`y = a(x - 1)^2 + 5`

The y intercept lets us solve for a


`2 = a(0 - 1)^2 + 5 = a+5`


`a=-3`

So our parabola is


`y = -3(x - 1)^2 + 5`


The x intercepts are the zeros of the polynomial. We've already completed the square, so let's solve directly for the zeros:


`0 = -3(x - 1)^2 + 5`


`3(x - 1)^2 = 5`


`(x - 1)^2 = \frac 5 3`


`x - 1 = \pm √(\frac5 3)`


`x = 1 \pm √(\frac53)`

So our x intercepts are


`(1- √(\frac53), 0) \quad \textrm{and} \quad (1+√(\frac53), 0)`

I hate ruining a nice exact answer with an approximation so I'll leave the calculator part to you.