Question 30·Easy·Advanced Math

The two forms must give the same result for every $y$. Type $y_1=[1...5]$; Desmos lists $1,2,3,4,5$. These inputs are nonzero, so the term containing $p$ won't disappear as it would at $y=0$. Desmos uses $y_1$ as data rather than as a graph variable.

Type $9y_1-4y_1^2\sim y_1(9-py_1)$. The $\sim$ tells Desmos to fit $p$ so the two forms agree across the list. Under PARAMETERS, it shows $p=4$. That makes the $-py_1^2$ term equal the original $-4y_1^2$ term, so the value of $p$ is $4$. Choice D.

No Desmos needed for this route. The common factor is visible. Both terms contain $y$, so pull out that shared factor. Dividing $9y$ by $y$ leaves $9$, and dividing $-4y^2$ by $y$ leaves $-4y$:

The given form is $y(9-py)$. For it to equal $y(9-4y)$ for every $y$, the coefficients of $y$ inside must match. Set them equal:

Multiply both sides by $-1$:

So the value of $p$ is $4$. Choice D.