|
Question 622103: Graph the quadratic equation by plotting points?
Graph the quadratic equation by plotting points
y=-2x^2+3
I found the vertex which is (0,3) so I graphed that point. What is the other point that I need to graph? I don't know how to get the other point in order to complete the graph of the parabola. Can you please include the other point and how you got the other point, thanks.
Found 2 solutions by jim_thompson5910, ewatrrr:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Plug in x = 1 (basically any x value you want) and follow the same method shown before
y=-2x^2+3
y=-2(1)^2+3
y=-2(1)+3
y=-2+3
y=1
So another point is (1,1)
By symmetry, another point is (-1,1) since x = -1 and x = 1 are both one unit away from the line of symmetry x = 0.
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website!
Hi,
y = -2x^2 + 3, Yes! V(0,3), opening downward as -2 < 0
y -axis the axis of symmetry, Pt(1,1) on the parabola.
Plot points and use the axis of symmetry to sketch the graph
Might recommend downloading the FREE graph software at http://www.padowan.dk.com
to check Your work when determining graphs of various functions You will be working with
|
|
|
| |
