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

Nov 30, 2017

The graph (the Parabola) Opens Up, as the coefficient of ${x}^{2}$ term is greater than ZERO.

Vertex $= \left(- 1 , - 2.5\right)$ and Axis of Symmetry $x = - 1$ Graph attached for a visual proof.

#### Explanation:

Standard Form of a Quadratic Equation is given by
$a {x}^{2} + b x + c = 0$

If $a > 0$ then the Parabola will "Open Up". If $a < 0$ then the parabola will "Open Down".

Since in our Quadratic Equation, the coefficient of the ${x}^{2}$ term is found to be $> 0$, the Parabola "Opens Upward".

Also observe that, if the Parabola "Opens Upward" then our Vertex will be a "Minimum" of the Quadratic Function.

We are given the Quadratic Function $y = \left(\frac{1}{2}\right) {x}^{2} + x - 2$

The expression $\left(- \frac{b}{2 a}\right)$ will give us the x-coordinate value of the Vertex.

To obtain the y-coordinate value of the "Vertex", substitute the value obtained from using the expression $\left(- \frac{b}{2 a}\right)$ in our original equation.

Hence, $y = \left(\frac{1}{2}\right) {\left(- 1\right)}^{2} + \left(- 1\right) - 2$ will help us to get the y-coordinate value of the Vertex.

After simplification, we get $y = - 2.5$

Hence our Vertex$= \left(- 1 , - 2.5\right)$

And our Axis of Symmetry $x = - 1$

Please refer to the attached graph for a visual proof of our solution.

I hope this explanation is helpful.