Question

The polynomial $f(x)$ has degree 3. If $f(-1) = 15$, $f(0)= 0$, $f(1) = -5$, and $f(2) = 12$, then what are the $x$-intercepts of the graph of $f$?

Answer 1

For future reference, this site uses `\(` and `\)` for in-line LaTeX wrapping, and `\[` and `\]` for new lines, as opposed to the usual `$...$` and `$$...$$`.

Answer 2

As for your question: the intermediate value theorem guarantees that at least one x-intercept is between \(x=1\) and \(x=2\). You're explicitly given another at \(x=0\).

Given that you know four different points, it's possible you're asked to find the explicit form of the polynomial. Call it \(f(x)=ax^3+bx^2+cx+d\). Plug in the given points, and set up a system of equations to solve for \(a,b,c,d\).

Answer 3

SithsAndGiggles I got ax^3+bx^2+dx as the equation, but is there a quick way to find a, b, or d? Do I actually have to guess and check?

Answer 4

@SithsAndGiggles

Answer 5

The problem at hand isn't too unwieldy, but if you're looking for a quick solution you might try using a calculator.
\[\begin{cases}f(-1)=15\\f(0)=0\\f(1)=-5\\f(2)=12\end{cases}~~\implies~~\begin{cases}15=-a+b-c+d\\
0=0a+0b+0c+d\\
-5=-125a+25b-5c+d\\
12=8a+4b+2c+d\end{cases}\]
The second equation gives you \(d=0\), so you're right about not having a constant term and the system is reduced to three equations and three unknowns.

Answer 6

Thanks, but can you tell me how you got the set of equations

Answer 7

Each given point \(f(x)=y\) gives you an output \(y\) associated with each input \(x\). If \(f(x)=ax^3+bx^2+cx+d\), then \(f(-1)=a(-1)^3+b(-1)^2+c(-1)+d=-a+b-c+d\), for instance.

Answer 8

Lmao, I guess I forgot that.

Answer 9

Thanks I think I got it now

Answer 10

yw