Find the polynomial equation represented with points at: (0,0) (3,0) and (4,3.2)
I found the equation of f(x)=.2(x+2)^2 (x-2)^3
This is wrong...can you please explain where my mistake is?

Question

Find the polynomial equation represented with points at: (0,0) (3,0) and (4,3.2)
I found the equation of f(x)=.2(x+2)^2 (x-2)^3
This is wrong...can you please explain where my mistake is?

anonymous · Answer

this is my work, f(x)=a(x+0)^2(x-3)^3
3.2=a(4+0)^2(4-3)^3
3.2=a(16)(1)
a=.2
f(x)=.2(x+2)^2(x-2)^3
where am I going wrong?

anonymous · Answer

you have 3 points, should be a quadratic polynomial

anonymous · Answer

y intercepts at 0 and 3 tells you it looks like $p(x)=ax(x-3)$ and solve for $a$ by taking $p(4)=a\times 4\times (4-3)=3.2$

anonymous · Answer

I still don't get it...

anonymous · Answer

you have three points 
a quadratic looks like $p(x)=ax^2+bx+c$ with three unknowns, namely $a,b,c$ so if you have three points, you are specifying a quadratic polynomial (degree two ) not some higher degree one
i am not sure where you got the exponent of 3 and 2 from in your answer

anonymous · Answer

the 3 and 2 are the double zero and triple zero

anonymous · Answer

you have the zeros are $0$ and $3$ which means it factors as $x(x-3)$ except the leading coefficient does not have to be 1, so it is 
$$p(x)=ax(x-3)$$

anonymous · Answer

does it say that 0 is a zero with multiplicity two?

anonymous · Answer

it didn't say that above, but if so, no matter, you would be correct and it would be 
$$f(x)=ax^2(x-3)^3$$

anonymous · Answer

no, it showed me the graph and I had to find the equation...
I would show you the graph, but I d'ont think that I can

anonymous · Answer

ok so lets imagine that it has a zero at 0 with multiplicity 2, (so it touches right?) and somehow you know that the zero at 3 has multiplicity 3
then your work is right, but i don't know how you changed the correct work to get a different answer

anonymous · Answer

$$f(4)=3.2=16a$$ so $a=0.2$ and your answer is 
$$f(x)=.2x^2(x-3)^3$$

anonymous · Answer

? lost

anonymous · Answer

you have the answer. you can multiply it out of you like, but if it factors as 
$f(x)=0.2x^2(x-3)^3$ it will not factor some other way

anonymous · Answer

you already have it in factored form. 
you can leave it like that, or you can multiply out
but you cannot multiply out, then factor again and get something other than what you started with

anonymous · Answer

ok, so pretend that there is a graph with points (-2,0) (1,0) and (-1,-4)
it would be f(x)=a(x+2)^2(x-1)^3
right?

anonymous · Answer

yes, assuming that for some reason you knew the multiplicity of the zeros
not sure how you know it though

anonymous · Answer

if I was looking at another graph...
then if than equation was right, then f(x)=2(x+2)^2(x-1)^3
and then I would get the same points...

anonymous · Answer

but I'm not...