# Question #f986a

Oct 4, 2017

$y = {x}^{2}$:
graph{x^2 [-40, 40, -20, 20]}

$- 6 + y = x$
graph{x+6 [-80, 80, -40, 40]}

#### Explanation:

Let's graph these one by one.

$y = {x}^{2}$ is your standard form of a parabola, and a pretty common shape.

You can graph this by subbing in points:
when $x = 1 , y = {\left(1\right)}^{2} = 1$
when $x = 2 , y = {\left(2\right)}^{2} = 4$
when $x = 3 , y = {\left(3\right)}^{2} = 9$ and so on.

Or, you can memorise what this graph looks like. It has a vertex at the origin, with two arms branching upwards on either side of the y-axis:
graph{x^2 [-40, 40, -20, 20]}

$- 6 + y = x$ is a different type of graph. By looking at it, you can tell it is line (i.e. a straight line) as its highest power of $x$ is 1.

(Note: $1 x$ is the same as $x$.)

To graph this one, you need to find the x- and y-intercepts:

x-intercept (when y=0):
$- 6 + 0 = x$
$x = - 6$

So the graph cuts the x-axis at $- 6$.

y-intercept (when x=0):
$- 6 + y = 0$
$y = 6$

So the graph cuts the y-axis at $6$.

You then plot these two points on a graph and draw a straight line that goes through both points:

graph{x+6 [-80, 80, -40, 40]}