Hello, magentarita!
Henderson is absolutely correct!
Did you make a sketch? . . . It's very simple.
A right triangle has one vertex on the graph of $\displaystyle y \:=\: 9  x^2,\;\;x > 0,\text{ at }(x, y),$
another at the origin, and the third on the positive xaxis at (x, 0).
Express the area $\displaystyle A$ of the triangle as a function of $\displaystyle x.$ Code:

9 **
 * P
 *(x,y)
 *: *
 *::: *
 *:::::
 +    +  *  
O Q 3
The area of a triangle is: .$\displaystyle A \;=\;\tfrac{1}{2}\text{(base)(height)} $
The base is $\displaystyle x.$
The height is $\displaystyle y$, where $\displaystyle y \:=\:9x^2$
Therefore: .$\displaystyle A \;=\;\tfrac{1}{2}x(9x^2)$