# What is the vertex form of y=41x^2-3x+17?

Nov 26, 2015

The wording 'vertex form' is new to me but I am assuming it is completion of the square:
$\textcolor{g r e e n}{y = 41 {\left(x - \frac{3}{82}\right)}^{2} + 16 \frac{155}{164}}$

#### Explanation:

If I am wrong about the term then perhaps I am showing you something else that may prove useful.

$\textcolor{b l u e}{S t e p 1}$
Write as $y = 41 \left({x}^{2} - \frac{3}{41} x\right) + 17. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

At the moment I can use the equals because I have not changed any of the total values on the right hand side (RHS). However, the next stage does change the value on the right so at that point I must not use the equals sign.
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$\textcolor{b l u e}{S t e p 2}$
Consider the $- \frac{3}{41} x$. Suppose we change this to $- \left(\frac{1}{2} \times \frac{3}{41}\right) x$
We get $\textcolor{w h i t e}{\ldots} \textcolor{b l u e}{\frac{3}{82}} \textcolor{red}{x} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(2\right)$

Suppose I rewrite (1) and use the $\textcolor{b l u e}{\frac{3}{82}}$ from (2)

$y \to 41 \left(x - \textcolor{b l u e}{\frac{3}{82}}\right)$

$\textcolor{g r e e n}{\text{ Notice that the second "x " has gone and the first is no longer squared.}}$

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$\textcolor{b l u e}{S t e p 3}$

Now square the bracket

$y \to 41 {\left(x - \frac{3}{82}\right)}^{\textcolor{b l u e}{2}}$
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$\textcolor{b l u e}{S t e p 4}$

Add the constant ( the right still does not $= y \textcolor{w h i t e}{. .}$ yet.

$y \to 41 {\left(x - \frac{3}{82}\right)}^{2} + 17. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(3\right)$
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$\textcolor{b l u e}{S t e p 5}$
The way we have written the bracket introduces an error. Let me multiply the bracket out and you will see what I mean.

$y \to 41 \left({x}^{2} - \frac{6}{82} x + \textcolor{b l u e}{\frac{9}{6724}}\right) + 17$

$y \to \left(41 {x}^{2} - \frac{41 \times 6}{82} x + \textcolor{b l u e}{\frac{41 \times 9}{6724}}\right) + 17$

$y \to 41 {x}^{2} - 3 x + \textcolor{b l u e}{\frac{9}{164}} + 17$

We have introduced the extra value of $\textcolor{b l u e}{\frac{9}{164}}$

To compensate for this we remove it by subtracting the same value

$\textcolor{g r e e n}{\underline{\text{NOW IT IS = } Y}}$

$y = \left(41 {x}^{2} - 3 x + \textcolor{b l u e}{\frac{9}{164}}\right) \textcolor{b r o w n}{- \frac{9}{164}} + 17$

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$\textcolor{b l u e}{S t e p 6}$

So the vertex form (completing the square) is:
Adjust equation (3) by subtracting $\textcolor{b r o w n}{- \frac{9}{164}}$

$y = 41 {\left(x - \frac{3}{82}\right)}^{2} \textcolor{b r o w n}{- \frac{9}{164}} + 17$

$\textcolor{g r e e n}{y = 41 {\left(x - \frac{3}{82}\right)}^{2} + 16 \frac{155}{164}}$

color(red)(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~)