How do you graph the equation by plotting points #y = -x^2 + 3#?

1 Answer
smendyka
Aug 8, 2017

See a solution process below:

Explanation:

First, we can write a table of values:

enter image source here

We can then plot the points as:

graph{((x+4)^2+(y+13)^2-0.125)((x+3)^2+(y+6)^2-0.125)((x+2)^2+(y+1)^2-0.125)((x+1)^2+(y-2)^2-0.125)((x+0)^2+(y-3)^2-0.125)((x-1)^2+(y-2)^2-0.125)((x-2)^2+(y+1)^2-0.125)((x-3)^2+(y+6)^2-0.125)((x-4)^2+(y+13)^2-0.125)=0 [-20, 20, -15, 5]}}

Now, we can draw a line through the points to graph the parabola:

graph{(y+(x^2)-3)((x+4)^2+(y+13)^2-0.125)((x+3)^2+(y+6)^2-0.125)((x+2)^2+(y+1)^2-0.125)((x+1)^2+(y-2)^2-0.125)((x+0)^2+(y-3)^2-0.125)((x-1)^2+(y-2)^2-0.125)((x-2)^2+(y+1)^2-0.125)((x-3)^2+(y+6)^2-0.125)((x-4)^2+(y+13)^2-0.125)=0 [-20, 20, -15, 5]}}

Related questions
See all questions in Graphs in the Coordinate Plane
Impact of this question
1781 views around the world
You can reuse this answer
Creative Commons License