# How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry given y=1/10x^2+x-2?

Jul 19, 2017

The graph 'opens upwards'.

Vertex$\to \left(x , y\right) = \left(- 5 , - \frac{9}{2}\right)$

Axis of symmetry $\to x = - 5$

#### Explanation:

$\textcolor{b l u e}{\text{Opens up or down}}$

If the coefficient of ${x}^{2}$ (the number in front of it :could be 1 )
is positive then the graph is of form $\cup$

If the coefficient of ${x}^{2}$ is negative then the graph is of form $\cap$

In this case the coefficient is $+ \frac{1}{10}$ so it is of form $\cup \to$ opens up.
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$\textcolor{b l u e}{\text{Determine the vertex}}$

There are several ways but I am going to show you a shortcut. The method forms part of the process for 'completing the square.

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

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

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

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

We have the equation:
$y = \frac{1}{10} {x}^{2} + x - 2$

Where $a = \frac{1}{10} \mathmr{and} \textcolor{red}{b = 1} \to \text{from } x = \textcolor{red}{1} x$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(1 \div \frac{1}{10}\right)$

${x}_{\text{vertex}} = - 5$

Substitute for $x = - 5$ to find ${y}_{\text{vertex}}$

${y}_{\text{vertex}} = \frac{1}{10} {\left(- 5\right)}^{2} + \left(- 5\right) - 2$

${y}_{\text{vertex}} = 2.5 - 5 - 2 = - 4.5 \to - \frac{9}{2}$

Vertex$\to \left(x , y\right) = \left(- 5 , - \frac{9}{2}\right)$
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$\textcolor{b l u e}{\text{Determine axis of symmetry}}$

This is a line that is parallel to the y-axis and passes through ${x}_{\text{vertex}}$

So axis of symmetry is $x = - 5$