Question

write a function in any form that would match the graph shown below.

Explanation:

Step1: Identify roots and multiplicity

The graph touches the x - axis at \(x=-1\) (wait, no, looking at the graph, it crosses at \(x = - 4\) and has a double root at \(x=-1\) (since it touches the x - axis there). Wait, actually, from the graph, the x - intercepts: one is at \(x=-4\) (crossing) and a double root at \(x=-1\) (touching). Let's assume the function is a cubic or quartic? Wait, the graph has a local maximum at \(x=-1\) (touching x - axis) and a root at \(x=-4\), and also when \(x = 0\), \(y=-6\). Let's try to write it in factored form.

If there is a root at \(x=-4\) (linear factor: \((x + 4)\)) and a double root at \(x=-1\) (quadratic factor: \((x + 1)^2\)), so the function could be \(f(x)=a(x + 4)(x + 1)^2\). Now we need to find \(a\).

Step2: Find the value of \(a\)

We know that when \(x = 0\), \(y=-6\). Substitute \(x = 0\) and \(y=-6\) into the function:

\(-6=a(0 + 4)(0 + 1)^2\)

\(-6=a\times4\times1\)

\(4a=-6\)

\(a=-\frac{6}{4}=-\frac{3}{2}\)

So the function is \(f(x)=-\frac{3}{2}(x + 4)(x + 1)^2\). Let's expand it to check:

First, expand \((x + 1)^2=x^{2}+2x + 1\)

Then multiply by \((x + 4)\): \((x + 4)(x^{2}+2x + 1)=x^{3}+2x^{2}+x+4x^{2}+8x + 4=x^{3}+6x^{2}+9x + 4\)

Then multiply by \(-\frac{3}{2}\): \(f(x)=-\frac{3}{2}x^{3}-9x^{2}-\frac{27}{2}x - 6\). Let's check the y - intercept: when \(x = 0\), \(f(0)=-6\), which matches the graph. Also, the end - behavior: as \(x
ightarrow-\infty\), the leading term \(-\frac{3}{2}x^{3}\) goes to \(+\infty\) (since odd degree and negative leading coefficient), and as \(x
ightarrow+\infty\), it goes to \(-\infty\), which matches the graph.

Answer:

\(f(x)=-\frac{3}{2}(x + 4)(x + 1)^2\) (or expanded form \(f(x)=-\frac{3}{2}x^{3}-9x^{2}-\frac{27}{2}x - 6\))