# How do you find the vertex of  y = 10x – 3x^2?

May 2, 2017

Using a 'sort of' cheat. Not really! It is part of the process of completing the square.

Vertex$\to \left(x , y\right) = \left(\frac{5}{3} , \frac{25}{3}\right)$

#### Explanation:

Note that as $- 3 {x}^{2}$ is negative the graph is of shape $\cap$

Write as $y = - 3 {x}^{2} + 10 x + 0$

This is now in the standardised form of

$y = a {x}^{2} + b x + c$

Change this to:

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

Note that $a \times \frac{b}{a} = b$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{10}{- 3} = + \frac{10}{6} = \frac{5}{3}$
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Substitute ${x}_{\text{vertex}} = + \frac{5}{3}$ giving:

${y}_{\text{vertex}} = - 3 {\left(\frac{5}{3}\right)}^{2} + 10 \left(\frac{5}{3}\right)$

${y}_{\text{vertex}} = - \frac{75}{9} + \frac{50}{3}$

${y}_{\text{vertex}} = - \frac{75}{9} + \frac{150}{9}$

${y}_{\text{vertex}} = + \frac{75}{9} = 8 \frac{1}{3} \to \frac{25}{3}$
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Vertex$\to \left(x , y\right) = \left(\frac{5}{3} , \frac{25}{3}\right)$ 