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

Dec 30, 2017

The vertex is $\left(\frac{7}{24} , - \frac{143}{48}\right)$.

#### Explanation:

First expand ${\left(3 x - 2\right)}^{2} = 9 {x}^{2} - 12 x + 4$.

Substituting that in we have:

$y = - 3 {x}^{2} - 5 x - \left(9 {x}^{2} - 12 x + 4\right)$

Distribute the negative:

$y = - 3 {x}^{2} - 5 x - 9 {x}^{2} + 12 x - 4$

Collect like terms:

$y = - 12 {x}^{2} + 7 x - 4$

The vertex is $\left(h , k\right)$ where $h = - \frac{b}{2 a}$ and $k$ is the value of $y$ when $h$ is substituted.

$h = - \frac{7}{2 \left(- 12\right)} = \frac{7}{24}$.

$k = - 12 {\left(\frac{7}{24}\right)}^{2} + 7 \left(\frac{7}{24}\right) - 4 = - \frac{143}{48}$ (I used a calculator...)

The vertex is $\left(\frac{7}{24} , - \frac{143}{48}\right)$.

Dec 30, 2017

$\left(\frac{7}{24} , - \frac{143}{48}\right)$

#### Explanation:

$\text{we require to express in standard form}$

$\Rightarrow y = - 3 {x}^{2} - 5 x - \left(9 {x}^{2} - 12 x + 4\right)$

$\textcolor{w h i t e}{\Rightarrow y} = - 3 {x}^{2} - 5 x - 9 {x}^{2} + 12 x - 4$

$\textcolor{w h i t e}{\Rightarrow y} = - 12 {x}^{2} + 7 x - 4 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{given the equation of a parabola in standard form then}$
$\text{the x-coordinate of the vertex is}$

${x}_{\textcolor{red}{\text{vertex}}} = - \frac{b}{2 a}$

$\text{here } a = - 12 , b = 7 , c = - 4$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{7}{- 24} = \frac{7}{24}$

$\text{substitute this value into the equation for y}$

$y = - 12 {\left(\frac{7}{24}\right)}^{2} + 7 \left(\frac{7}{24}\right) - 4 = - \frac{143}{48}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{7}{24} , - \frac{143}{48}\right)$