# How do you find the vertex and intercepts for 4x - y² + 6y - 1 = 0?

May 15, 2016

color(blue)(y_("intercepts")~~5.828" and "0.172" to 3 decimal places")
$\textcolor{b l u e}{{x}_{\text{intercept}} = \frac{1}{4}}$
color(blue)("Vertex "->(x,y)->(-2,3)

#### Explanation:

The structure of this is such that it implies $f \left(y\right)$ instead of $f \left(x\right)$ so that is how I am going to treat it. The consequence is that it rotates an $f \left(x\right)$ graph so that the axis of symmetry is parallel to the x-axis instead of the y-axis.

Write as:
$\text{ } 4 x = {y}^{2} - 6 y + 1$

Divide both sides by 4

$\text{ } x = \frac{1}{4} {y}^{2} - \frac{6}{4} y + \frac{1}{4}$

$\textcolor{b l u e}{\text{Determine the vertex}}$

Write as:

$\text{ } x = \frac{1}{4} \left({y}^{2} - 6 y\right) + \frac{1}{4}$

${y}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(- 6\right) = + 3$

x_("vertex")=1/4(3)^2-3/2(3)+1/4" "=" "-2

color(blue)("Vertex "->(x,y)->(-2,3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine "x" intercept}}$

Set $y = 0$

$\textcolor{b l u e}{\implies {x}_{\text{intercept}} = \frac{1}{4} {\left(0\right)}^{2} - \frac{3}{2} \left(0\right) + \frac{1}{4}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine "y" intercept}}$

Set $x = 0$

$\implies \frac{1}{4} \left({y}^{2} - 6 y\right) + \frac{1}{4} = 0$

$\textcolor{b r o w n}{\text{Completing the square}}$

$\frac{1}{4} {\left(y - 3\right)}^{2} + \frac{1}{4} + k = 0$

Where $k$ is the constant of correction.

$k = \left(- 1\right) \times \left[\frac{1}{4} {\left(- 3\right)}^{2}\right] = - \frac{9}{4}$ giving:

color(brown)(1/4(y-3)^2+1/4 +k=0)color(blue)(" "->" "1/4(y-3)^2-2=0
'....................................................................

$\implies {\left(y - 3\right)}^{2} = 8$

$y - 3 = \pm \sqrt{8} \text{ } = \pm 2 \sqrt{2}$

${y}_{\text{intercept}} = 3 \pm 2 \sqrt{2}$

color(blue)(y_("intercepts")~~5.828" and "0.172" to 3 decimal places")
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~