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

Mar 2, 2017

The x-intercepts are given by

$x = 5 \pm \frac{4 \sqrt{3}}{3}$

#### Explanation:

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

$y = a {\left(x - p\right)}^{2} + q$

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

$\frac{13}{4} = a {\left(6 - 5\right)}^{2} + 4$

$\frac{13}{4} = a {\left(1\right)}^{2} + 4$

$- \frac{3}{4} = a$

The equation is therefore

$y = - \frac{3}{4} {\left(x - 5\right)}^{2} + 4$

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

$0 = - \frac{3}{4} {\left(x - 5\right)}^{2} + 4$

$- 4 = - \frac{3}{4} {\left(x - 5\right)}^{2}$

$- \frac{4}{- \frac{3}{4}} = {\left(x - 5\right)}^{2}$

$\frac{16}{3} = {\left(x - 5\right)}^{2}$

$\pm \frac{4}{\sqrt{3}} = x - 5$

$x = 5 \pm \frac{4}{\sqrt{3}}$

$x = 5 \pm \frac{4 \sqrt{3}}{3}$

A graphical depiction of the parabola confirms our findings.

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

Hopefully this helps!