Help with writing equation for cubic polynomial?

[elis94camaro]

Ok so im stuck on this and dont know how to approach it. It isnt like the other typical zeros root problems ive seen where they give you the x intercepts.Here it is Write an equation for a cubic polynomial P(x)with leading coefficient −1 whose graph passes through the point (2, 8) and is tangent to the x axis at the origin.All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got the right answer from the choices but i did it by plugging in the point (2,8) and it came out to be -x^2(x-4) or -X^3+4x^2. Please help me as I have no clue on how to go about solving thanks.

 

[Someone2841]

Ok so im stuck on this and dont know how to approach it. It isnt like the other typical zeros root problems ive seen where they give you the x intercepts.Here it is Write an equation for a cubic polynomial P(x)with leading coefficient −1 whose graph passes through the point (2, 8) and is tangent to the x axis at the origin.All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got the right answer from the choices but i did it by plugging in the point (2,8) and it came out to be -x^2(x-4) or -X^3+4x^2. Please help me as I have no clue on how to go about solving thanks.

The best way to solve a problem is to first list the conditions mathematically:

$\displaystyle \text{1) } P(x) = -x^3 + a x^2+b x+c$

$\displaystyle \text{2) } P(2) = 8$

$\displaystyle \text{3) } P'(0) = 0$

$\displaystyle \text{4) } P(0) = 0$


From $\displaystyle \text{2)}$ we get $\displaystyle -8 + 4a + 2b + c = 8$

From $\displaystyle \text{3)}$ we get $\displaystyle -3(0)^2 + 2a(0) + b = 0$ and therefore $\displaystyle b = 0$ and because of that $\displaystyle 4a + c = 16$.


Now we can plug this into $\displaystyle \text{1)}$ to get: $\displaystyle P(x) = -x^3 + a x^2 - 4a + 16$ and from $\displaystyle \text{4)}$ solve for a:

$\displaystyle P(0) = -4a + 16 = 0 \to a = 4$


Plug this back into $\displaystyle \text{1)}$ to get: $\displaystyle P(x) = -x^3 + 4 x^2$

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You can also go the easy way out by making this assumptions:

1) Any polynomial will be tangent at point (a,0) if and only if the multiplicity of a as a root of the polynomial is greater than 1.

Then we know the polynomial will be in this form:

$\displaystyle P(x) = -x^2(x+c)$

Now solve for $\displaystyle c$ in $\displaystyle P(2) = 8$

$\displaystyle 8 = -4(2+c) \to c = -4$

And we get our same polynomial $\displaystyle P(x) = -x^2(x-4) = -x^3 + 4 x^2$

 

[elis94camaro]

Thanks but i dont understand where you got the equation from number 3? Did you take the derivative from the original equation of -x^3+ax^2+bx+ c and get that?Could you also elaborate on when it is appropriate to use this rule 1) Any polynomial will be tangent at point (a,0) if and only if the multiplicity of a as a root of the polynomial is greater than 1. and how you figured the multiplicity from the info in the question? Sorry for asking these questions but i just didnt understand this problem, it was the only one on the homework that stumped me.

 

[burakaltr]

Ok so im stuck on this and dont know how to approach it. It isnt like the other typical zeros root problems ive seen where they give you the x intercepts.Here it is Write an equation for a cubic polynomial P(x)with leading coefficient −1 whose graph passes through the point (2, 8) and is tangent to the x axis at the origin.All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got the right answer from the choices but i did it by plugging in the point (2,8) and it came out to be -x^2(x-4) or -X^3+4x^2. Please help me as I have no clue on how to go about solving thanks.

Tangent to origin at (0,0) is equivalent to say that the function has duble roots. zero and zero

Rephrasing f(x) = A(x-0)(x-0)(x-a)
f(x)=A x**3 - a A x**2

A=-1

- x**3 +a x**2

point ( 2, 8 )

-8+ 4a =8
a=4

f(x) therefore is = - x**3 + 4 x**2

 

[elis94camaro]

Thanks for the help, but the thing is that this problem was a homework problem in my college precal class and we havent touched on anything about derivative and calculus yet, so that why i didnt approach it using calculus.

 

[Deleted member 4993]

Thanks for the help, but the thing is that this problem was a homework problem in my college precal class and we havent touched on anything about derivative and calculus yet, so that why i didnt approach it using calculus.

If read the previous explanations carefully, you'll find that this can be explained considering the fact that the function has a tangent at the origin.