# How do you graph x^2 = y-x?

Sep 8, 2017

$\text{see explanation}$

#### Explanation:

$\text{the parabola can be graphed by finding the zeros, vertex}$
$\text{and additional points}$

$\text{express in standard form } y = a {x}^{2} + b x + c$

$\Rightarrow y = {x}^{2} + x$

$\text{with } a = 1 , b = 1 , c = 0$

$\text{since "c=0" then y-intercept is 0}$

• " if "a>0" then minimum turning point " uuu

• " if "a<0" then maximum turning point " nnn

$\text{here "a=1>0rArr" minimum turning point}$

$\text{to find the zeros let y = 0 and solve for x}$

$\Rightarrow {x}^{2} + x = 0$

$\Rightarrow x \left(x + 1\right) = 0$

$\Rightarrow x = 0 \text{ or "x=-1larrcolor(red)" zeros}$

$\text{find the vertex by "color(blue)"completing the square}$

$\Rightarrow {x}^{2} + 2 \left(\frac{1}{2}\right) x + \frac{1}{4} - \frac{1}{4}$

$= {\left(x + \frac{1}{2}\right)}^{2} - \frac{1}{4} \leftarrow \textcolor{red}{\text{ in vertex form}}$

$\text{the vertex } = \left(- \frac{1}{2} , - \frac{1}{4}\right)$

$\textcolor{b l u e}{\text{Additional points}}$

$x = 1 \to y = {1}^{2} + 1 = 2 \Rightarrow \left(1 , 2\right)$

$x = 2 \to y = {2}^{2} + 2 = 6 \Rightarrow \left(2 , 6\right)$

$x = - 2 \to y = {\left(- 2\right)}^{2} - 2 = 2 \Rightarrow \left(- 2 , 2\right)$

$\text{plot these key points and draw a smooth curve through them}$
graph{x^2+x [-10, 10, -5, 5]}