# What is the vertex of  y= 4(x+2)^2-x^2-5x+3?

Nov 9, 2016

The coordinate of the vertex is $\left(- \frac{11}{6} , \frac{107}{12}\right)$.

#### Explanation:

For the parabola given by the standard-form equation $y = a {x}^{2} + b x + c$, the $x$-coordinate of the parabola's vertex is at $x = - \frac{b}{2 a}$.

So, to find the vertex's $x$-coordinate, we should first write the equation of this parabola in standard form. To do so, we have to expand ${\left(x + 2\right)}^{2}$. Recall that ${\left(x + 2\right)}^{2} = \left(x + 2\right) \left(x + 2\right)$, which can then be FOILed:

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

$\textcolor{w h i t e}{y} = 4 \left({x}^{2} + 4 x + 4\right) - {x}^{2} - 5 x + 3$

Distribute the $4$:

$\textcolor{w h i t e}{y} = 4 {x}^{2} + 16 x + 16 - {x}^{2} - 5 x + 3$

Group like terms:

$\textcolor{w h i t e}{y} = \left(4 {x}^{2} - {x}^{2}\right) + \left(16 x - 5 x\right) + \left(16 + 3\right)$

$\textcolor{w h i t e}{y} = 3 {x}^{2} + 11 x + 19$

This is now in standard form, $y = a {x}^{2} + b x + c$. We see that $a = 3 , b = 11$, and $c = 19$.

So, the $x$-coordinate of the vertex is $x = - \frac{b}{2 a} = - \frac{11}{2 \left(3\right)} = - \frac{11}{6}$.

To find the $y$-coordinate, plug $x = - \frac{11}{6}$ into the parabola's equation.

$y = 3 {\left(- \frac{11}{6}\right)}^{2} + 11 \left(- \frac{11}{6}\right) + 19$

$\textcolor{w h i t e}{y} = 3 \left(\frac{121}{36}\right) - \frac{121}{6} + 19$

$\textcolor{w h i t e}{y} = \frac{121}{12} - \frac{121}{6} + 19$

$\textcolor{w h i t e}{y} = \frac{121}{12} - \frac{242}{12} + \frac{228}{12}$

$\textcolor{w h i t e}{y} = \frac{107}{12}$

So, the coordinate of the vertex is $\left(- \frac{11}{6} , \frac{107}{12}\right)$.

graph{4(x+2)^2-x^2-5x+3 [-33.27, 31.68, -5.92, 26.56]}

Note that $\left(- \frac{11}{6} , \frac{107}{12}\right) \approx \left(- 1.83 , 8.92\right)$.