I am stumbling on an (at first sight) simple homework example. Maybe someone can help me. I just would need tips how to build up the equations for the respective polynomials, not the full solution.

Here is the homework example: An even polynomial of fourth degree $f(x)$ has a zero point at $(-2,0)$ and on another zero point it has a tangent with the equation $y = 6 - 6 \cdot x$. The polynomial of second degree $g(x)$ is intersecting with $f(x)$ at a maximum of $f(x)$, furthermore $g(x)$ itself has a maximum at $(0.5,5.25)$.

The task is now to determine the polynomials $f(x)$ and $g(x)$, so to determine their coefficients. So I have to determine the three coefficients of $f(x) = a_1 \cdot x^4 + a_2 \cdot x^2 + a_3$ and the three coefficients of $g(x) = b_1 \cdot x^2 + b_2 \cdot x + b_3$.

So how to get the needed three equations for determining $a_1,a_2,a_3$ and the three equations for determing $b_1,b_2,b_3$ ?


Do it one step at a time.

$f(-2)=0$ gives you one equation.

If $f$ has $y=6-6x$ as a tangent at one of its zero crossings, then the zero crossings in question must be at the point where $y=6-6x$ crosses the $x$ axis. Setting $f(x)=0$ there gives you another equation. Finally, the slope of $f$ at that point must be $-6$, so setting $f'(x)=-6$ at that point gives you a third equation. Solve. Now you know $f$.

Alternatively, you know one zero of $f$; and another zero where the $6-6x=0$. Since $f$ is even, you can reflect those two zeroes in the $y$-axis to get two more, and then you have all the four possible zeroes of a fourth-degree polynomial. So $f(x)=c(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ for some $c$ -- but it turns out that setting $c=1$ already gives you the right slope at the point of tangency. How nice!

Now that you know $f$ you can sketch it and find that it has exactly one local maximum, which must be where it crosses $g$. That gives you one equation for the coefficients of $g$; knowing that $g(0.5)=5.25$ and $g'(0.5)=0$ gives you two more. Solve again!

Alternatively since you know the apex of $g$, you can be sure that $g(x)=d(x-0.5)^2+5.25$ for some factor $d$; then you just need to find the right $d$ such that $g$ intesects $f$ at the maximum, and multiply out.

(The alternatives at each step may seem more involved, but at least for me they're actually the simpler way. This way around I can keep everything in my head, up to and including finding $g(x)=-5x^2+5x+4$ -- whereas I wouldn't be able to do the three-equations-in-three-unknowns route without taking it to pencil and paper).


First condition (given): $f(-2) = 0$.

Second condition (from evenness, may not be useful!): $f(2) = 0$.

Third condition: $f'(x_0) = 6$.

Fourth condition: $f(x_0) = 6-6x_0$.

Fifth condition: At some $x_1$, $f'(x_1) = 0$ and $g(x_1) = f(x_1)$.

Sixth condition: $g'(0.5) = 0, g(0.5) = 5.25$

Seventh condition (since $g$ has a maximum it must open downwards): $b_1 < 0$.

Can you use these to find what you need?

  • $\begingroup$ @HenningMakholm That's what I intended. I meant that $x_1$ is a single specific $x_1$. But that may not be clear, let me edit! Thanks :) $\endgroup$ – Emily Jan 9 '15 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.