Question

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

Answer 1

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]} :