# What is the vertex form of y=x^2+3x+2?

Jan 19, 2016

(-3/2;-1/4)

#### Explanation:

The vertex or turning point occurs at the point when the derivative of the function (slope) is zero.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 0 \iff 2 x + 3 = 0$

$\iff x = - \frac{3}{2}$.

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

$= - \frac{1}{4}$.

Thus the vertex or turning point occurs at (-3/2;-1/4).

The graph of the function verifies this fact.

graph{x^2+3x+2 [-10.54, 9.46, -2.245, 7.755]}

Jan 19, 2016

$\textcolor{g r e e n}{\text{Vertex Form } \textcolor{w h i t e}{\ldots} \to} \textcolor{w h i t e}{\ldots} \textcolor{b l u e}{y = {\left(x + \frac{3}{2}\right)}^{2} - \frac{1}{4}}$

#### Explanation:

Given: $\textcolor{w h i t e}{\ldots .} y = {x}^{2} + 3 x + 2$.....................(1)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider just the ${x}^{2} + 3 x$

We are going to convert this to a 'perfect square' that is not quite equal to it. We then apply a mathematical 'adjustment' such that becomes equal to it.

$\textcolor{b r o w n}{\text{Step 1}}$

Change the ${x}^{2} \text{ to just } x$
Change the $3 \text{ in "3x" to } \frac{1}{2} \times 3 = \frac{3}{2}$

Put it together in the form of ${\left(x + \frac{3}{2}\right)}^{2}$

As yet ${\left(x + \frac{3}{2}\right)}^{2}$ does not equal ${x}^{2} + 2 x$ so we need to find out how to adjust it.

The adjustment is $\left({x}^{2} + 2 x\right) - {\left(x + \frac{3}{2}\right)}^{2}$

$\left({x}^{2} + 2 x\right) - \left({x}^{2} + 3 x + \frac{9}{4}\right)$

So the adjustment is $- \frac{9}{4}$

$\textcolor{b r o w n}{\text{Note that the "+9/4" is an introduced value that is not wanted} .}$ $\textcolor{b r o w n}{\text{So we have to remove it; hence } - \frac{9}{4}}$

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

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Step 2}}$

Substitute (2) into equation (1) giving:

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

$\textcolor{g r e e n}{\text{Vertex Form } \textcolor{w h i t e}{\ldots} \to} \textcolor{w h i t e}{\ldots} \textcolor{b l u e}{y = {\left(x + \frac{3}{2}\right)}^{2} - \frac{1}{4}}$