How do you use the important points to sketch the graph of #y=-1/5x^2#?

1 Answer
Yoshie · Stefan V.
Jun 11, 2018

Below.

Explanation:

Clearly, this graph is just a transformation of the parabola #y = x^2#

graph{x^2 [-5, 5, -1, 10]}

The first thing to note is that there is a negative before #x^2# so it is an inverse parabola

graph{-x^2 [-5, 5, -10, 1]}

Then substitute #y = 0# to get

#0 = -(1/5)x^2#

Divide both sides by #1/5#

#0 = -x^2#

#sqrt0 = -x#

#0 = -x#

#x = 0#

So the first major point #= (0,0)#

Now substitute #x = 1#

#y = -(1/5)1^2#

#y = -(1/5)1#

#y = -(1/5)#

The second important point #= (1, -1/5)#

Now substitute #x = 2#

#y = -(1/5)2^2#

#y = -(1/5)4#

#y = -(4/5)#

The third important point #= (2, -4/5)#

Now you have 3 points: #(0,0), (1, -1/5),# and #(2, -4/5)# and we know it's an inverse parabola so draw these points on a graph and then use them to draw an inverse parabola. And you get

graph{-(1/5)x^2 [-5, 5, -10, 1]}

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
1683 views around the world
You can reuse this answer
Creative Commons License