\( \def\dfrac#1#2{\displaystyle\frac{#1}{#2}} \def\solve#1{\begin{array}{rcl}#1\end{array} } \)

Home / 05 Polynomial Functions / 05 Equation Of Polynomial From Graph

Example: Use the graph below to determine the equation of the polynomial. Use \(k\) for the leading coefficient, \(-k\) if it should be negative.


Solution: The process for determining an equation for a polynomial can be laid out in a series of steps:

  1. We need to determine the zeros of the polynomial.
  2. We need to determine the (minimum) multiplicity of each zero.
  3. Using this, we write out each of the factors.
  4. To determine the sign of the leading coefficient, we look at the orientation and minimum degree of the polynomial.
  5. To determine the actual coefficient value, we need at least one point that is not a zero. If it is not provided, we cannot determine it. In this case, we have been instructed to just use \(k\).

So, let's write down the zeros in order: -4, 0, and 5. To determine the multiplicity, we examine the graph in detail near each of the zeros:

By looking closely at how the function crosses the horizontal axis, we can see that the factor near \(x=-4\) behaves linearly, the factor near \(x=0\) behaves like a cubic, and the factor near \(x=5\) behaves like a quadratic.Thus, we can state that the minimum degree of each factor reflects that behavior and write the following:

\[ f(x) = Ax^3(x+4)(x-5)^2 \]

Note that each factor has the exponent that corresponds to the behavior we observed. To determine the sign of the leading coefficient, first we get the degree: 3+1+2 = 6. Then we observe the end behavior (both go down) so we can state that it is Even degree with negative leading coefficient!

\[ f(x) = -kx^3(x+4)(x-5)^2 \]