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The diagram below shows a curve with equation of the form ${y = kx(x + a)^2}$, which passes through the points (-2, 0), (0, 0) and (1, 3).

enter image description here

What are the values of a and k.

I know my roots are x = -2, x = 0 and x = 3.

But as the y intercept is 0, I don't see how I can get any meaningful values for k:

${k(x + 2)(x - 3) = 0}$

How can I find the values of k and a?

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    $\begingroup$ $x=3$ is not a root $\endgroup$ – angryavian Jan 9 '16 at 0:24
  • $\begingroup$ Expand the expression and sub in the coordinates to get a equation set. Solve it to get the value of $a$ and $k$. $\endgroup$ – user122049 Jan 9 '16 at 0:29
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Be careful, $x=3$ is not a root. The curve does not cross the $x$-axis when $x=3$. There are only two roots: $x=-2$ and $x=0$.

You know that $(-2,0)$, $(0,0)$ and $(1,3)$ are all points on the curve.

You should substitute $x=-2$ and $y=0$ to get one equation. Then substitute $x=0$ and $y=0$ to get a second equation. Finally substitute $x=1$ and $y=3$ to get a third equation. Then solve these simultaneously.

As you rightly say: $(x,y)=(0,0)$ doesn't give you anything.

When we substitute $(x,y)=(-2,0)$ into $y=kx(x+a)^2$ we get $0=-2k(-2+a)^2$. So, either $k=0$ or $a=2$. Obviously $k \neq 0$ or the whole question would collapse from $y=kx(x+a)^2$ down to $y=0$ which can't be right because the original graph isn't the line $y=0$. That means $a=2$.

When we substitute $(x,y)=(1,3)$ into $y=kx(x+a)^2$ we get $3=k(1+a)^2$. We know that $a=2$, so this becomes $3=k(1+2)^2$, which in turn becomes $3=9k$. This tells us that $k=\frac{1}{3}$.

Given that $a=2$ and $k=\frac{1}{3}$, the equation $y=kx(x+a)^2$ becomes $$\color{red}{y=\frac{1}{3}x(x-2)^2}$$

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Observe that there is a double root at $-2$ since the curve is tangent to the x-axis and a single root at the origin. To find $k$ plug in the point $(1,3)$ and solve for $k$.

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x= -2 is a double root so, since you are told that it is a polynomial must have $(x- (-2))^2$ as a factor. x= 0 is a root so f(0)= 0. f(1)= 3.

Further, this is clearly not a quadratic! You are told that the function is of the form "$kx(x- a)^2$" not $k(x- 2)(x- 3)$. The fact that x= -1 is a double root tells you what a is. Then $k(1)(1- a)^2= 3$.

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$$ y = kx(x + a)^2 $$

Using that the curve includes $P=(-2,0)$ and $Q=(1,3)$ we have $$ 0 = -2 k (-2 +a)^2 \\ 3 = k(1 + a)^2 $$ We note that $k \ne 0$, otherwise the curve would have been constant zero. So the first equation turns into $$ 0 = (a - 2)^2 $$ which gives $a = 2$. This turns the second equation into $$ 3 = k (1 +2)^2 = 9 k \iff k = 1/3 $$ Reproduced Graph:

reproduced graph

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  • $\begingroup$ what did you use to draw the graph? $\endgroup$ – dagda1 Jan 9 '16 at 7:26
  • $\begingroup$ GeoGebra. A very nice software. It is written in Java and except for the 3D graphics part very portable. I use the ports for Windows, Linux and iOS iPad. $\endgroup$ – mvw Jan 9 '16 at 7:29

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