# How do you graph f(x)=x^2?

Aug 1, 2015

This is a vertical parabola - a sort of U shape - with vertex at $\left(0 , 0\right)$, axis of symmetry $x = 0$, passing through $\left(4 , - 2\right)$, $\left(1 , - 1\right)$, $\left(1 , 1\right)$ and $\left(2 , 4\right)$.

#### Explanation:

Ultimately, to graph (almost) any function $f \left(x\right)$ you can compute $f \left(x\right)$ for several values of $x$ to find some points $\left(x , f \left(x\right)\right)$ through which the graph passes.

In our case $f \left(- 2\right) = {\left(- 2\right)}^{2} = 4$ gives us $\left(- 2 , 4\right)$, $f \left(- 1\right) = 1$ gives us $\left(- 1 , 1\right)$, etc.

In the general case of quadratic functions of form $f \left(x\right) = a {x}^{2} + b x + c$ you can reformulate to find the vertex, axis of symmetry and where the parabola intersects the axes.

For example,

$f \left(x\right) = a {x}^{2} + b x + c = a {\left(x - \left(- \frac{b}{2 a}\right)\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

$= a {\left(x - h\right)}^{2} + k$

with $h = - \frac{b}{2 a}$ and $k = c - {b}^{2} / \left(4 a\right)$

This is in vertex form: The vertex is at $\left(h , k\right)$.

In our particular example, $a = 1$, $b = c = 0$, so these formulae simplify to give $\left(h , k\right) = \left(0 , 0\right)$

graph{(y-x^2)((x+2)^2+(y-4)^2-0.02)((x+1)^2+(y-1)^2-0.02)(x^2+y^2-0.02)((x-1)^2+(y-1)^2-0.02)((x-2)^2+(y-4)^2-0.02)(y*0.00001+x) = 0 [-10.5, 9.5, -2.28, 7.72]}