How do you graph #x= -3(y-5)^2 +2#?

1 Answer
Alexander L.
Aug 9, 2015

The graph is an "n" shape . From the equations alone, we can tell for sure that this is a quadratic equation (either "u" or "n" shaped).

Explanation:

Expand the equation to get;
#x=-3y^2+30y-73#

  • Find turning points and determine if they are maximum points or minimum points.
  • Next, find points of intersection on the vertical and horizontal axis.

Finding turning points (#df(x)/dx=0 #);

#dx/dy=-6y+30# where #dx/dy=0#

Hence, #y=5#. When #y=5, x=2#
The turning point is at coordinate #(5,2)# and it is a maximum point since the graph is an "n" shape. You can tell the "n" shape if the coefficient for the #y^2# is a negative .

Finding intersections :

Vertical axis ;
Let #y=0#,
#x-3(0-5)^2+2#.
#x=-73#

Horizontal axis :
Use #(-b+-sqrt(b^2-4ac))/(2a).#

You should get something like this (scroll on the graph to get a better view):

PS: Feel free to ask away any questions.

graph{-3x^2+30x-73 [-11.25, 11.25, -5.625, 5.625]} :

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