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

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;
$x = - 3 {y}^{2} + 30 y - 73$

• 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 ($\mathrm{df} \frac{x}{\mathrm{dx}} = 0$);

$\frac{\mathrm{dx}}{\mathrm{dy}} = - 6 y + 30$ where $\frac{\mathrm{dx}}{\mathrm{dy}} = 0$

Hence, $y = 5$. When $y = 5 , x = 2$
The turning point is at coordinate $\left(5 , 2\right)$ 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 $y = 0$,
$x - 3 {\left(0 - 5\right)}^{2} + 2$.
$x = - 73$

Horizontal axis :
Use $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} .$

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]} :