# How do you find the important points to graph f(x)=2x^2?

Jul 10, 2018

See explanation

#### Explanation:

Set $y = f \left(x\right) = 2 {x}^{2}$

Compare to the standardised form: $y = a {x}^{2} + b x + c$
This equation $\textcolor{w h i t e}{\text{dddddfdd") ->color(white)("ddddd}} y = 2 {x}^{2} + 0 x + 0$

Key points:

$a > 0 \to \left(\text{positive}\right)$ so the graph is of form $\cup$

$c \to$ y-intercept $= 0$

${x}_{\text{vertex") -> (-1/2)xx b/acolor(white)("ddd") =color(white)("ddd") (-1/2)xx0/2color(white)("d")=color(white)("d}} 0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As ${x}_{\text{vertex}} = 0$ the axis of symmetry is the y-axis

As $a > 0$ then the general shape is $\cup$ with the y-axis in the middle.

Couple this with $c = 0$ and it means that the vertex is at $\left(x , y\right) = \left(0.0\right)$