How do you graph #y= -1/4x^2# by plotting points?

1 Answer
Aug 21, 2017

Answer:

See a solution process below:

Explanation:

First, we can plot these points for the equation:

enter image source here

graph{((x-8)^2+(y+16)^2-0.3)((x-6)^2+(y+9)^2-0.3)(x^2+y^2-0.3)((x+6)^2+(y+9)^2-0.3)((x+8)^2+(y+16)^2-0.3)=0 [-30, 30, -20, 10]}

Now, we can draw the parabola through the points:

graph{(y+1/4x^2)((x-8)^2+(y+16)^2-0.3)((x-6)^2+(y+9)^2-0.3)(x^2+y^2-0.3)((x+6)^2+(y+9)^2-0.3)((x+8)^2+(y+16)^2-0.3)=0 [-30, 30, -20, 10]}