check workings for polynomial function from graph The diagram shows a curve with equation of the form $y = kx(x + a)^2$,
which passes through $(-2, 0)$, $(0, 0)$ and $(1, 3)$.
What are the values of $a$ and $k$.


To find $k$, I would set $f(x) = 1$.
$f(1) = k(x + 2)(x + 0)(x - 1)$
$3 = k(1 + 2)(x + 1)(1 - 1)$
$3 = k(3)(1)(0)$
This would make $k = 0$ which does not appear to be right.
$$y=kx(x+a)^2\quad \quad \quad (-2, 0),(0, 0) , (1, 3)$$


$$\text{for} (0,0):\quad \quad \quad f(0)=0=0\\
\text{for}(1,3):\quad \quad \quad f(1)=k(1+a)^2=3\\
\text{for}(-2,0):\quad \quad \quad f(-2)=-2k(-2+a)^2=0$$

so you have two equations with two variables:

$$\begin{cases}k(1+2a+a^2)=3\\
-2k(4-4a+a^2)=0\end{cases}$$

$$\begin{cases}k+2ak+ka^2=3\\
-8k+8ka-2ka^2=0\end{cases}$$

Solve, you should get $\color{red}{a=2,k=\frac 1 3}$
The equation is satisfied by $(-2,0)$ and $(1,3)$
Putting $(-2,0)$ in the equation we obtain:
$$0 = k(-2)(-2+a)^2 $$
Therefore, $(a-2)=0$ as $k \ne 0$
$$\implies a=2$$
Similarly, putting $(1,3)$ in the equation we obtain:
$$3 = k*1*(1+2)$$
Therefore $k=1$