# How do you find the vertex and intercepts for x – 4y^2 + 16y – 19 = 0?

Jul 26, 2018

Vertex $\left(3 , 2\right)$
X-intercept $\left(19 , 0\right)$

#### Explanation:

$x - 4 {y}^{2} + 16 y - 19 = 0$

$x - 19 = 4 {y}^{2} - 16 y$

$x - 19 + 16 = 4 \left({y}^{2} - 4 y + 4\right)$

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

${\left(y - 2\right)}^{2} = \frac{1}{4} \left(x - 3\right)$ which is in the form ${\left(y - k\right)}^{2} = 4 a \left(x - h\right)$ where $\left(h , k\right)$ is the vertex

Hence $\left(3 , 2\right)$ is the vertex.

For the intercepts,
When $y = 0$
${\left(0 - 2\right)}^{2} = \frac{1}{4} \left(x - 3\right)$
$4 = \frac{1}{4} \left(x - 3\right)$
$16 = x - 3$
$x = 19$

When $x = 0$,
${\left(y - 2\right)}^{2} = \frac{1}{4} \left(0 - 3\right)$
${\left(y - 2\right)}^{2} = \frac{1}{4} \left(- 3\right)$
${\left(y - 2\right)}^{2} = - \frac{3}{4}$
Since you cannot squareroot a negative number, there are no y-intercepts

Below is what the graph looks like

graph{x-4y^2+16y-19=0 [-10, 10, -5, 5]}

Jul 26, 2018

#### Explanation:

The equation is

$x - 4 {y}^{2} + 16 y - 19 = 0$

$4 {y}^{2} - 16 y = x - 19$

${y}^{2} - 4 y = \frac{1}{4} \left(x - 19\right)$

Completing the square

${y}^{2} - 4 y + 4 = \frac{1}{4} \left(x - 19\right) + 4$

Factorising

${\left(y - 2\right)}^{2} = \frac{1}{4} x - \frac{19}{4} + 4 = \frac{1}{4} x - \frac{3}{4}$

The equation of the parabola is

${\left(y - 2\right)}^{2} = \frac{1}{4} \left(x - 3\right)$

Comparing this to the equation of a parabola

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

The vertex is $V = \left(a , b\right) = \left(3 , 2\right)$

The intercepts are when $y = 0$

$\implies$, $4 = \frac{1}{4} \left(x - 3\right)$

$\implies$, $16 = x - 3$

$\implies$, $x = 16 + 3 = 19$

The point is $= \left(19 , 0\right)$

graph{x-4y^2+16y-19=0 [0.74, 20.74, -3.32, 6.68]}