Question

the function $f(x)$ is shown on the coordinate plane below.

graph of $y = f(x)$

which of the following is the equation of the function $f(x)$ graphed above?

a $f(x) = x(x - 2)^2(x - 1)(x + 1)$

b $f(x) = - (x - 2)^2(x - 1)(x + 1)$

c $f(x) = - x(x - 2)^2(x - 1)(x + 1)$

d $f(x) = (x - 2)^2(x - 1)(x + 1)^2$

Explanation:

Step1: Identify x-intercepts

The graph intersects the x - axis at \(x=- 1\), \(x = 0\), \(x = 1\), and \(x=2\). At \(x = 2\), the graph touches the x - axis and turns around, so the multiplicity of the root \(x = 2\) is 2. At \(x=-1\), \(x = 0\), and \(x = 1\), the graph crosses the x - axis, so their multiplicities are 1. So the factored form of the function should have factors \((x + 1)\), \(x\), \((x - 1)\), and \((x - 2)^2\).

Step2: Determine the leading coefficient sign

As \(x\to+\infty\), the graph of the function goes to \(-\infty\). For a polynomial function \(y=a(x - r_1)^{m_1}(x - r_2)^{m_2}\cdots(x - r_n)^{m_n}\), the end - behavior is determined by the leading term \(ax^{m_1 + m_2+\cdots+m_n}\). The degree of our polynomial (sum of multiplicities) is \(1 + 1+1 + 2=5\) (odd degree). For an odd - degree polynomial, if as \(x\to+\infty\), \(y\to-\infty\), the leading coefficient \(a\) is negative.

Looking at the options:

• Option A: \(f(x)=x(x - 2)^2(x - 1)(x + 1)\), leading coefficient is positive (product of positive coefficients of the linear factors), so as \(x\to+\infty\), \(y\to+\infty\) (since degree 5 is odd), which does not match the graph.
• Option B: \(f(x)=-(x - 2)^2(x - 1)(x + 1)\), this function does not have a factor of \(x\) (since the graph passes through \(x = 0\)), so it is incorrect.
• Option C: \(f(x)=-x(x - 2)^2(x - 1)(x + 1)\), it has factors \(x\) (root at \(x = 0\)), \((x + 1)\) (root at \(x=-1\)), \((x - 1)\) (root at \(x = 1\)), \((x - 2)^2\) (root at \(x = 2\) with multiplicity 2) and the leading coefficient is negative. The degree is \(1+2 + 1+1=5\) (odd), and as \(x\to+\infty\), \(y=(-1)x^{5}+\cdots\), so as \(x\to+\infty\), \(y\to-\infty\) which matches the end - behavior of the graph.
• Option D: \(f(x)=(x - 2)^2(x - 1)(x + 1)^2\), it has a factor \((x + 1)^2\) (multiplicity 2 for \(x=-1\)) and no factor of \(x\) (since the graph passes through \(x = 0\)), so it is incorrect.

Answer:

C. \(f(x)=-x(x - 2)^2(x - 1)(x + 1)\)