# How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry given y=5x^2+1?

Dec 9, 2017

$y = 5 {x}^{2} + 1$ opens up, the vertex is (0,1), and the axis of symmetry is on the y axis (equation is x=0)

#### Explanation:

Up or Down? To tell whether a quadratic equation's graph opens up or down, look at whether the coefficient of x squared (number before x squared) is greater than or less than 0 (it can't be zero, cause then it isn't a quadratic equation). For example, $y = 5 {x}^{2} + 1$ has 5, so it is positive and opens up. If it is $y = - 5 {x}^{2} + 1$, then it has -5, which is negative and opens down.

Vertex? If you are using this structure of quadratic equations ($y = a {x}^{2} + b x + c$ ) which you are, then you plug b and a of that formula to find the x coordinate of the vertex: $- \frac{b}{2} a$. So since b is 0 (cause your equation has no __x, just x squared and a number 1) and a is 5, you plug it in:
$x = - \frac{b}{2} a$
$x = - \frac{0}{2} \left(5\right)$
$x = - \frac{0}{10}$
$x = 0$

So you have x=0 for the vertex. To find the y coordinate, just plug in x=0 into the equation:
$y = 5 {x}^{2} + 1$
$y = 5 {\left(0\right)}^{2} + 1$
$y = 0 + 1$
$y = 1$

So now you have the coordinates for the vertex of $y = 5 {x}^{2} + 1$, which is $\left(0 , 1\right)$

Axis of Symmetry? This is easy once you find the vertex. Because the vertex is sitting on the axis of symmetry, you just take the x value for up-down quadratic graphs (not sideways, that would be totally different) which, in this case, is $0$, and set x to always = $0$ and there you have it, you got the equation for the axis of symmetry - $x = 0$

A really helpful tool to have a visual guide is to use this graphing calculator, desmos.com/calculator where you can plug in any equation you want and have it graph it for you. Good luck on math!