Question

graph shown below?
answer
\\( f(x) = (x - 8)(x - 5)^2 \\)
\\( f(x) = (x - 8)^2(x - 5) \\)
\\( f(x) = (x + 8)(x + 5)^2 \\)
\\( f(x) = (x + 8)^2(x + 5) \\)

Explanation:

Step1: Identify x-intercepts

The graph crosses or touches the x - axis at \(x = 5\) and \(x=8\). At \(x = 5\), the graph crosses the x - axis, so the multiplicity of the root at \(x = 5\) is odd. At \(x = 8\), the graph touches the x - axis and turns around, so the multiplicity of the root at \(x = 8\) is even.

Step2: Write the function from roots

If a polynomial has a root at \(x=a\), then \((x - a)\) is a factor of the polynomial. For a root with odd multiplicity (crossing the axis), the factor has an odd exponent, and for a root with even multiplicity (touching the axis), the factor has an even exponent.
Since the root at \(x = 5\) has odd multiplicity and the root at \(x = 8\) has even multiplicity, the polynomial should be of the form \(f(x)=(x - 8)^{2}(x - 5)\) (because \((x - 8)\) corresponds to the root \(x = 8\) with even multiplicity and \((x - 5)\) corresponds to the root \(x = 5\) with odd multiplicity). We can also check the end - behavior: the leading term of \(f(x)=(x - 8)^{2}(x - 5)=(x^{2}-16x + 64)(x - 5)=x^{3}-21x^{2}+144x - 320\). The leading coefficient is positive and the degree is 3 (odd), so as \(x
ightarrow+\infty\), \(f(x)
ightarrow+\infty\) and as \(x
ightarrow-\infty\), \(f(x)
ightarrow-\infty\), which matches the end - behavior of the given graph.

Answer:

\(f(x)=(x - 8)^{2}(x - 5)\)