How do you graph $y = 2 x^{2} - 8 x - 1$ $y=2x^2-8x-1$ by plotting points?

Question

3329 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-05T16:17:04

Make a table of $x$ $x$ and $y$ $y$ .

Try plugging in an $x$ $x$ -value: a good one to start with is always $0$ $0$ :

$y = 2 {(0)}^{2} - 8 (0) - 1$ $y=2(0)^2-8(0)-1$
$y = 0 - 0 - 1$ $y=0-0-1$
$y = - 1$ $y=-1$

Thus, when $x = 0$ $x=0$ , $y = - 1$ $y=-1$ , so we can plot the point $(0, - 1)$ $(0,-1)$ on our graph.

Continue to find points and graph them:
enter image source here

Eventually, you should see the shape of the graph starting to form.

Connect the dots, and you should see this:

graph{2x^2-8x-1 [-26.94, 24.38, -11.3, 14.36]}

How do you graph y=2x^2-8x-1y=2x2−8x−1y=2x^2-8x-1 by plotting points?