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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.

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  • $\begingroup$ Don't you mean $f(1) = 3$? $\endgroup$ – Pichi Wuana Mar 6 '16 at 19:19
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$$y=kx(x+a)^2\quad \quad \quad (-2, 0),(0, 0) , (1, 3)$$

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

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 $\color{red}{a=2,k=\frac 1 3}$

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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 $k=1$

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