# What is the vertex form of y = -3x^2+9x+1?

Dec 6, 2015

$y = - 3 {\left(x - \frac{3}{2}\right)}^{2} + \frac{31}{4}$

#### Explanation:

Given:$\textcolor{w h i t e}{. .} y = - 3 {x}^{2} + 9 x + 1. \ldots \ldots \ldots . \left(1\right)$

Write as:$\textcolor{w h i t e}{. .} y = - 3 \left({x}^{2} \textcolor{g r e e n}{- 3 x}\right) + 1$

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Consider the RHS only

Write as: $- 3 {\left(x - \frac{3}{2}\right)}^{2} + 1. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$

The $\left(- \frac{3}{2}\right)$ comes from halving the coefficient of $x \text{ in } \textcolor{g r e e n}{- 3 x}$

Expression (2) has an inherent error which we need to correct

$- 3 {\left(x - \frac{3}{2}\right)}^{2}$
$= - 3 \left({x}^{2} - 3 x + \frac{9}{4}\right)$
$= - 3 {x}^{2} + 9 x - \frac{27}{4.} \ldots \ldots \ldots \ldots \ldots \ldots \left(3\right)$

Add the constant of +1 as shown in equation (1) giving

$= - 3 {x}^{2} + 9 x - \frac{27}{4} + 1. \ldots \ldots \ldots \ldots \ldots \ldots \left({3}_{a}\right)$

When you compare $\left({3}_{a}\right)$ to (1) you see that the error introduced is $- \frac{27}{4}$

We correct for this by removing it from the vertex form equations using $\textcolor{b l u e}{+ \frac{27}{4}}$

Thus the $\underline{\textcolor{red}{\text{incorrect}}}$ form of $y = - 3 {\left(x - \frac{3}{2}\right)}^{2} + 1 \textcolor{b l u e}{\text{ is adjusted by:}}$

$y = - 3 {\left(x - \frac{3}{2}\right)}^{2} + 1 \textcolor{b l u e}{+ \frac{27}{4}}$

Giving:

$y = - 3 {\left(x - \frac{3}{2}\right)}^{2} \textcolor{b r o w n}{+ \frac{31}{4}}$