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 \(x=-5\) (wait, no, looking at the graph, the x - intercepts where it touches (has a double root) are at \(x = - 5\)? Wait, no, the graph touches the x - axis at \(x=-5\)? Wait, no, the graph has a double root at \(x=-5\)? Wait, no, looking at the graph, the left - most touch point is at \(x=-5\)? Wait, no, the graph touches the x - axis at \(x=-5\)? Wait, no, the graph has a double root at \(x=-5\) and \(x = 2\)? Wait, no, the graph touches the x - axis at \(x=-5\) (wait, the grid: from the graph, the left touch point is at \(x=-5\)? Wait, no, the x - axis crossings: the graph touches the x - axis at \(x=-5\) (wait, the x - coordinate of the left touch point: between - 6 and - 4, let's see, the vertex is at \(x=-5\)? Wait, no, the graph has a double root at \(x=-5\) and \(x = 2\)? Wait, no, the graph touches the x - axis at \(x=-5\) (a double root) and \(x = 2\) (a double root). So the function has factors of \((x + 5)^2\) and \((x - 2)^2\). So the function is of the form \(y=a(x + 5)^2(x - 2)^2\).

Step2: Find the value of \(a\)

We can use the y - intercept. The y - intercept is at \((0,-40)\). Substitute \(x = 0\) and \(y=-40\) into the function:
\(-40=a(0 + 5)^2(0 - 2)^2\)
First, calculate \((5)^2=25\) and \((-2)^2 = 4\). Then \(25\times4 = 100\). So the equation becomes \(-40=a\times100\). Solve for \(a\): \(a=\frac{-40}{100}=-\frac{2}{5}\).

Step3: Write the function

Substitute \(a =-\frac{2}{5}\) into the factored form: \(y =-\frac{2}{5}(x + 5)^2(x - 2)^2\). We can expand this or leave it in factored form. Let's expand it:
First, \((x + 5)^2=x^{2}+10x + 25\) and \((x - 2)^2=x^{2}-4x + 4\)
Multiply \((x^{2}+10x + 25)(x^{2}-4x + 4)=x^{4}-4x^{3}+4x^{2}+10x^{3}-40x^{2}+40x + 25x^{2}-100x + 100=x^{4}+6x^{3}-11x^{2}-60x + 100\)
Then multiply by \(-\frac{2}{5}\): \(y=-\frac{2}{5}x^{4}-\frac{12}{5}x^{3}+\frac{22}{5}x^{2}+24x - 40\)
Alternatively, we can also consider the general form of a quartic function with double roots at \(x=-5\) and \(x = 2\). Another way: the function is a quartic (degree 4) with leading coefficient negative (since the ends go down).

So the function can be written as \(y =-\frac{2}{5}(x + 5)^2(x - 2)^2\) or expanded form.

Answer:

\(y =-\frac{2}{5}(x + 5)^2(x - 2)^2\) (or equivalent expanded form)