# What is the vertex form of y=2x^2+11x+12?

Mar 12, 2017

Yhe vertex form is $y = 2 {\left(x + \frac{11}{4}\right)}^{2} - \frac{25}{8}$

#### Explanation:

To find the vertex form, you complete the square

$y = 2 {x}^{2} + 11 x + 12$

$y = 2 \left({x}^{2} + \frac{11}{2} x\right) + 12$

$y = 2 \left({x}^{2} + \frac{11}{2} x + \frac{121}{16}\right) + 12 - \frac{121}{8}$

$y = 2 {\left(x + \frac{11}{4}\right)}^{2} - \frac{25}{8}$

The vertex is $= \left(- \frac{11}{4} , - \frac{25}{8}\right)$

The symmetry line is $x = - \frac{11}{4}$

graph{(y-(2x^2+11x+12))(y-1000(x+11/4))=0 [-9.7, 2.79, -4.665, 1.58]}

Mar 12, 2017

$\textcolor{b l u e}{y = 2 {\left(x + \frac{11}{4}\right)}^{2} - \frac{25}{8}}$

#### Explanation:

Consider the standardised form of $y = a {x}^{2} + b x + c$

The vertex form is: $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k + c$

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$\textcolor{b r o w n}{\text{Additional note about the method}}$
By rewriting the equation in this form you introduce an error. Let me explain.

Multiply out the bracket in $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c$ and you get:

$y = a \left[{x}^{2} + \frac{2 x b}{2 a} + {\left(\frac{b}{2 a}\right)}^{2}\right] + c$

$\textcolor{g r e e n}{y = a {x}^{2} + b x + \textcolor{red}{a {\left(\frac{b}{2 a}\right)}^{2}} + c}$

the $\textcolor{red}{a {\left(\frac{b}{2 a}\right)}^{2}}$ is not in the original equation so it is the error. Thus we need to 'get rid' of it. By introducing the correction factor of $k$ and setting $\textcolor{red}{a {\left(\frac{b}{2 a}\right)}^{2} + k = 0}$ we 'force' the vertex form back into the value of the original equation.
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Given:$\text{ "y=ax^2+bx+c" "->" } y = 2 {x}^{2} + 11 x + 12$

$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k + c \text{ "->" } y = 2 {\left(x + \frac{11}{4}\right)}^{2} + k + 12$

But:
$a {\left(\frac{b}{2 a}\right)}^{2} + k = 0 \text{ "->" } 2 {\left(\frac{11}{4}\right)}^{2} + k = 0$

$\implies k = - \frac{121}{8}$

So by substitution we have:
$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k + c \text{ } \to y = 2 {\left(x + \frac{11}{4}\right)}^{2} - \frac{121}{8} + 12$

$\textcolor{b l u e}{y = 2 {\left(x + \frac{11}{4}\right)}^{2} - \frac{25}{8}}$
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The two equation have been plotted to show that they produce the same curve. One is thicker than the other so that they can both be seen.