# Thread: Write the equation of this cubic

1. ## Write the equation of this cubic

Ok so I need help finding out how you write cubic expressions just from looking at the graph.

The problem is that it's a strange cubic that cuts through the y-axis at 6.
I'm not sure how to ask this question but all I can give you is the answer of the graph in it's expression.

$y=-0.3(x-2)^2(x-5)$

So the curve it shows in the paper starts from the point 6 on the y-axis and then comes down to 2 on the x-axis where it touches the x-axis and then goes up and down again through 5 at the x-axis.
The only thing that boggles me is the 0.3, the one number that acts on the whole equation causing it to cut through the y-axis at 6.

It also states that the length of the object is 6m

That's the most detail I can provide into this question. Help would be awesome, thank you.

2. Originally Posted by Mooki
Ok so I need help finding out how you write cubic expressions just from looking at the graph.

The problem is that it's a strange cubic that cuts through the y-axis at 6.
I'm not sure how to ask this question but all I can give you is the answer of the graph in it's expression.

$y=-0.3(x-2)^2(x-5)$

So the curve it shows in the paper starts from the point 6 on the y-axis and then comes down to 2 on the x-axis where it touches the x-axis and then goes up and down again through 5 at the x-axis.
The only thing that boggles me is the 0.3, the one number that acts on the whole equation causing it to cut through the y-axis at 6.

It also states that the length of the object is 6m

That's the most detail I can provide into this question. Help would be awesome, thank you.
The model is clearly $y = a(x-2)^2 (x - 5)$. Now substitute the given point (0, 6) and solve for $a$.

3. Hello, Mooki!

From the graph, write the equation of the cubic.
Code:
        |
6 o
|
|*            *
| *        *    *
|   *     *       *
----+-------o----------o----
|       2          5
|
$f(x)$ has $x$-intercepts at 2 and 5.
. . Then $(x-2)$ and $(x-5)$ are factors of $f(x).$

Since the graph is tangent to the $x$-axis at $x = 2$
. . the factor $(x-2)$ is raised to an even power.

So, the function has the form: . $f(x) \:=\:a(x-2)^2(x-5)$ .[1]

Since $(0,6)$ is on the graph, we have: . $x = 0,\:y = 6$

Substitute into [1]: . $6 \:=\:a(0-2)^2(0-5) \quad\Rightarrow\quad 6 \:=\:a(\text{-}2)^2(5) \quad\Rightarrow\quad -20a \:=\:6$
. . Hence: . $a \:=\:-\frac{6}{20} \:=\:-0.3$

Therefore: . $f(x) \;=\;-0.3(x-2)^2(x-5)$

4. , here is the graph.

I think this one is what you wish to find

5. Thanks to everyone that helped out with the problem.

I see now that I was very close to solving it in the first place, just one thing off : P

GUYS R GRATE > : D YIAAAA