Question

write a function in any form that would match the graph shown below.
(graph: a coordinate plane with x-axis from -10 to 10 and y-axis from -20 to 20. the graph is a polynomial with roots at x=1 (double root), x=6 (double root), and y-intercept at (0,12).)
answer
attempt 1 out of 2
f(x) = blank
submit answer
you have up to 6 questions left to raise your score.

Explanation:

Step1: Identify the roots of the function

From the graph, we can see that the function touches the x - axis at \(x = 1\) (with a double root, since the graph touches the axis and turns around) and \(x=6\) (also a double root). So the roots of the function are \(x = 1\) (multiplicity 2) and \(x = 6\) (multiplicity 2). In factored form, a polynomial function with roots \(r_1,r_2,\cdots,r_n\) and leading coefficient \(a\) is given by \(f(x)=a(x - r_1)^{m_1}(x - r_2)^{m_2}\cdots(x - r_n)^{m_n}\), where \(m_i\) is the multiplicity of the root \(r_i\). So, we can write the function as \(f(x)=a(x - 1)^{2}(x - 6)^{2}\).

Step2: Determine the leading coefficient \(a\)

We know that the graph passes through the point \((0,12)\) (the y - intercept). Substitute \(x = 0\) and \(f(x)=12\) into the function \(f(x)=a(x - 1)^{2}(x - 6)^{2}\):

$$
LATEXBLOCK0
$$

Step3: Write the function

Substitute \(a=\frac{1}{3}\) into the factored form of the function. So the function is \(f(x)=\frac{1}{3}(x - 1)^{2}(x - 6)^{2}\). We can also expand this function:

First, expand \((x - 1)^{2}=x^{2}-2x + 1\) and \((x - 6)^{2}=x^{2}-12x + 36\)

Then, multiply \((x^{2}-2x + 1)(x^{2}-12x + 36)\):

$$
LATEXBLOCK1
$$

Then multiply by \(\frac{1}{3}\):

\(f(x)=\frac{1}{3}x^{4}-\frac{14}{3}x^{3}+\frac{61}{3}x^{2}-28x + 12\)

Answer:

\(f(x)=\frac{1}{3}(x - 1)^{2}(x - 6)^{2}\) (or the expanded form \(\frac{1}{3}x^{4}-\frac{14}{3}x^{3}+\frac{61}{3}x^{2}-28x + 12\))