Question

A parabola has a vertex at `(5, 4)` and passes through `(6, 13/4)`. What are the x-intercepts?

Answer 1

The x-intercepts are given by

`x = 5 +- (4sqrt(3))/3`

Explanation:

Start by finding the equation of the parabola. The vertex form of a parabola, with vertex `(p, q)`, is given by

`y = a(x- p)^2 + q`

We know an x-value, a y-value and the vertex. We can therefore set up an equation and solve for `a`.

`13/4 = a(6 - 5)^2 + 4`

`13/4 = a(1)^2 + 4`

`-3/4 = a`

The equation is therefore

`y = -3/4(x - 5)^2 + 4`

We can solve for the x-intercepts by taking the square root. Set `y` to `0`.

`0 = -3/4(x - 5)^2 + 4`

`-4 = -3/4(x - 5)^2`

`-4/(-3/4) = (x - 5)^2`

`16/3 = (x - 5)^2`

`+-4/sqrt(3) = x - 5`

`x = 5 +- 4/sqrt(3)`

`x = 5 +- (4sqrt(3))/3`

A graphical depiction of the parabola confirms our findings.

graph{y = -3/4(x - 5)^2 + 4 [-10, 10, -5, 5]}

Hopefully this helps!