# What are the vertex, focus and directrix of  y=3x^2+8x+17 ?

Jan 1, 2018

Vertex $\textcolor{b l u e}{= \left[- \frac{8}{6} , \frac{35}{3}\right]}$

Focus $\textcolor{b l u e}{= \left[- \frac{8}{6} , \frac{35}{3} + \frac{1}{12}\right]}$

Directrix $\textcolor{b l u e}{y = \left[\frac{35}{3} - \frac{1}{12}\right] \mathmr{and} y = 11.58333}$

Labelled Graph is also available

#### Explanation:

$\textcolor{red}{y = 3 {x}^{2} + 8 x + 17}$

Coefficient of the ${x}^{2}$ term is greater than Zero

Hence, our Parabola Opens Up and we will also have a Vertical Axis of Symmetry

We need bring our quadratic function to the form given below:

$\textcolor{g r e e n}{4 P \left(y - k\right) = {\left(x - h\right)}^{2}}$

Consider

$y = 3 {x}^{2} + 8 x + 17$

Note that, we need to keep both the $\textcolor{red}{{x}^{2}}$ and the $\textcolor{red}{x}$ term on one side and keep both the $\textcolor{g r e e n}{y}$ and the constant term on the other side.

To find the Vertex, we will Complete the Square on x

$\Rightarrow y - 17 = 3 {x}^{2} + 8 x$

Divide every single term by $3$ to get

$\Rightarrow \frac{y}{3} - \frac{17}{3} = \left(\frac{3}{3}\right) {x}^{2} + \left(\frac{8}{3}\right) x$

$\Rightarrow \frac{y}{3} - \frac{17}{3} = {x}^{2} + \left(\frac{8}{3}\right) x$

$\Rightarrow \frac{y}{3} - \frac{17}{3} + \textcolor{b l u e}{\square} = {x}^{2} + \left(\frac{8}{3}\right) x + \textcolor{b l u e}{\square}$

What value goes into the $\textcolor{b l u e}{B l u e \square}$?

Divide the coefficient of the x.term by $2$ and Square.

The answer goes into the $\textcolor{b l u e}{B l u e \square}$.

$\Rightarrow \frac{y}{3} - \frac{17}{3} + \textcolor{b l u e}{\frac{16}{9}} = {x}^{2} + \left(\frac{8}{3}\right) x + \textcolor{b l u e}{\frac{16}{9}}$

$\Rightarrow \left(\frac{1}{3}\right) y - \frac{17}{3} + \left(\frac{16}{9}\right) = {x}^{2} + \left(\frac{8}{3}\right) x + \left(\frac{16}{9}\right)$

$\Rightarrow \left(\frac{1}{3}\right) y - \frac{51 + 16}{9} = {x}^{2} + \left(\frac{8}{3}\right) x + \left(\frac{16}{9}\right)$

$\Rightarrow \left(\frac{1}{3}\right) y - \frac{35}{9} = {x}^{2} + \left(\frac{8}{3}\right) x + \left(\frac{16}{9}\right)$

$\Rightarrow \left(\frac{1}{3}\right) y - \frac{35}{9} = {\left[x + \left(\frac{8}{6}\right)\right]}^{2}$

Factor $\frac{1}{3}$ out on the Left-hand Side (LHS) to get

$\Rightarrow \left(\frac{1}{3}\right) \left[y - \frac{35}{3}\right] = {\left[x + \left(\frac{8}{6}\right)\right]}^{2}$

We can rewrite to bring it to the required form given below:

$\textcolor{g r e e n}{4 P \left(y - k\right) = {\left(x - h\right)}^{2}}$

$\Rightarrow \left(\frac{1}{3}\right) \left[y - \frac{35}{3}\right] = {\left[x - \left(- \frac{8}{6}\right)\right]}^{2}$

whered

$4 P = \frac{1}{3}$

$k = \frac{35}{3}$

$h = - \frac{8}{6}$

Hence, our Vertex will be

Vertex $\left(h , k\right) = \left\{\begin{matrix}- \frac{8}{6} \\ \frac{35}{3}\end{matrix}\right\}$

Using $4 P = \frac{1}{3}$, we get

$P = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$

Hence, $P = \frac{1}{12}$

Focus is always on the Axis of Symmetry

Focus is also inside the Parabola

Focus will have the same x.Value as the Vertex because it lies on the Axis of Symmetry

The Axis of Symmetry is at $x = - \frac{8}{6}$

The Directrix is always Perpendicular to the Axis of Symmetry

The Value of P tells us how far the Focus is from the Vertex

The Value of P also tells us how far the Directrix is from the Vertex

Since we know that $P = \frac{1}{12}$, Focus is $\frac{1}{12}$ or $0.83333$ units away from the Vertex

Our Focus is also $0.83333$ units away from the Vertex and lies on the Axis of Symmetry

Also, Focus is inside our parabola.

So, the Location of the Focus is given by

Focus $\textcolor{b l u e}{= \left[- \frac{8}{6} , \frac{35}{3} + \frac{1}{12}\right]}$

Directrix is always Perpendicular to the Axis of Symmetry

$\textcolor{b l u e}{y = \left[\frac{35}{3} - \frac{1}{12}\right] \mathmr{and} y = 11.58333}$ is the required equation of the Directrix and also lies on the Axis of Symmetry

Please refer to the graph below:

A labelled graph given below with a few intermediate calculations shows on it might also be useful