MHB Zero's question at Yahoo Answers regarding polynomial fitting

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The discussion focuses on constructing a fourth-degree polynomial with specified x-intercepts at -12, -6, and 6, and a y-intercept of -432. The polynomial must have a double root at x=6 due to its quartic nature, leading to the form f(x) = -k(x+12)(x+6)(x-6)², where k is a positive constant. By substituting the y-intercept condition, k is determined to be 1/6. The fully expanded polynomial is f(x) = -1/6x⁴ - x³ + 18x² + 36x - 432. This solution effectively meets the problem's requirements.
MarkFL
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Here is the question:

Write a function for the polynomial that fits the following description.?

write a function for the polynomial that fits the following description.

p is a fourth-degree polynomial with x-intercepts -12, -6, and 6 and y-intercept -432;
p(x) is positive only on the interval (-12 , -6 ).

Here is a link to the question:

Write a function for the polynomial that fits the following description.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Zero,

From the given information, we know the polynomial must have a graph resembling the following:

View attachment 629

Since the function does not pass through the $x$-axis at $x=6$, and it is a quartic, meaning it can have at most 4 roots, we know this root is of multiplicity 2.

Now, in order to have a negative $y$-intercept, we know it must have the form:

$f(x)=-k(x+12)(x+6)(x-6)^2$ where $0<k$

To determine $k$, we may use the information regarding its $y$ intercept as follows:

$f(0)=-k(0+12)(0+6)(0-6)^2=-432$

$-12\cdot6^3k=-432=-2\cdot6^3$

$\displaystyle k=\frac{1}{6}$

Hence:

$\displaystyle f(x)=-\frac{1}{6}(x+12)(x+6)(x-6)^2$

Since we are asked for a polynomial, we should expand it fully to find:

$\displaystyle f(x)=-\frac{1}{6}(x+12)(x-6)(x+6)(x-6)=$

$\displaystyle -\frac{1}{6}(x^2+6x-72)(x^2-36)=$

$\displaystyle -\frac{1}{6}(x^4-36x^2+6x^3-216x-72x^2+2592)=$

$\displaystyle -\frac{1}{6}(x^4+6x^3-108x^2-216x+2592)=$

$\displaystyle -\frac{1}{6}x^4-x^3+18x^2+36x-432$
 

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MarkFL said:
Hello Zero,

From the given information, we know the polynomial must have a graph resembling the following:

View attachment 629

Since the function does not pass through the $x$-axis at $x=6$, and it is a quartic, meaning it can have at most 4 roots, we know this root is of multiplicity 2.

Now, in order to have a negative $y$-intercept, we know it must have the form:

$f(x)=-k(x+12)(x+6)(x-6)^2$ where $0<k$

To determine $k$, we may use the information regarding its $y$ intercept as follows:

$f(0)=-k(0+12)(0+6)(0-6)^2=-432$

$-12\cdot6^3k=-432=-2\cdot6^3$

$\displaystyle k=\frac{1}{6}$

Hence:

$\displaystyle f(x)=-\frac{1}{6}(x+12)(x+6)(x-6)^2$

Since we are asked for a polynomial, we should expand it fully to find:

$\displaystyle f(x)=-\frac{1}{6}(x+12)(x-6)(x+6)(x-6)=$

$\displaystyle -\frac{1}{6}(x^2+6x-72)(x^2-36)=$

$\displaystyle -\frac{1}{6}(x^4-36x^2+6x^3-216x-72x^2+2592)=$

$\displaystyle -\frac{1}{6}(x^4+6x^3-108x^2-216x+2592)=$

$\displaystyle -\frac{1}{6}x^4-x^3+18x^2+36x-432$

I really dig your ability to present solutions in a clear and concise manner
 
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