Question

which equation choice could represent the graph shown below?
answer
$f(x) = x(x^2 + 36)$
$f(x) = (x + 6)(x^2 - 36)$
$f(x) = x(x^2 - 36)$
$f(x) = (x - 6)(x^2 - 36)$

Explanation:

Step1: Find x-intercepts from graph

The graph intersects the x - axis at \(x=-6\), \(x = 0\), and \(x=6\). So the roots of the function are \(x=-6\), \(x = 0\), and \(x = 6\).

Step2: Write the function in factored form

If the roots of a polynomial function \(f(x)\) are \(r_1\), \(r_2\), \(r_3\), then the function can be written as \(f(x)=a(x - r_1)(x - r_2)(x - r_3)\) (where \(a\) is a non - zero constant). Here, \(r_1=-6\), \(r_2 = 0\), \(r_3=6\), so \(f(x)=a(x+6)(x - 0)(x - 6)=a x(x + 6)(x - 6)\). We know that \((x + 6)(x - 6)=x^{2}-36\), so \(f(x)=a x(x^{2}-36)\). For the given graph, when we expand the options, if we take \(a = 1\), the function is \(f(x)=x(x^{2}-36)\) (since \((x + 6)(x - 6)=x^{2}-36\) and the root at \(x = 0\) gives the factor \(x\)).
Let's check other options:

• For \(f(x)=x(x^{2}+36)\), the roots of \(x^{2}+36 = 0\) are imaginary (\(x=\pm6i\)), so the graph will not intersect the x - axis at real points other than \(x = 0\), which does not match the graph.
• For \(f(x)=(x + 6)(x^{2}-36)=(x + 6)(x + 6)(x - 6)=(x + 6)^{2}(x - 6)\), the root \(x=-6\) has a multiplicity of 2, so the graph will touch the x - axis at \(x=-6\) (not cross it), which does not match the given graph.
• For \(f(x)=(x - 6)(x^{2}-36)=(x - 6)(x + 6)(x - 6)=(x - 6)^{2}(x + 6)\), the root \(x = 6\) has a multiplicity of 2, so the graph will touch the x - axis at \(x = 6\) (not cross it), which does not match the given graph.

Answer:

\(f(x)=x(x^{2}-36)\)