Question

write a function in any form that would match the graph shown below.
answer
attempt 1 out of 2
f(x) =

Explanation:

Step1: Identify the roots

The graph touches the x - axis at \(x=-5\) (wait, no, looking at the graph, the x - intercepts where the graph touches the x - axis (since it's a tangent, so double roots) are at \(x = - 5\)? Wait, no, the graph has a vertex at \(x=-5\)? Wait, no, looking at the x - axis, the points where the graph touches the x - axis are \(x=-5\)? Wait, no, the x - axis crossings (touching) are at \(x=-5\)? Wait, no, the graph: let's look at the x - axis. The first "touch" is at \(x=-5\)? Wait, no, the x - coordinates where the graph touches the x - axis (so the roots with multiplicity 2) are \(x=-5\) and \(x = 4\)? Wait, no, looking at the graph, the left vertex is at \(x=-5\) (since it's between - 6 and - 4, maybe \(x=-5\)) and the right vertex is at \(x = 4\)? Wait, no, the x - intercepts (where the graph touches the x - axis) are at \(x=-5\) and \(x = 4\)? Wait, no, let's re - examine. The graph has a "bump" on the left that touches the x - axis at \(x=-5\) (mid - point of - 6 and - 4) and a bump on the right that touches the x - axis at \(x = 4\) (mid - point of 2 and 6). Wait, no, the x - intercepts (touching) are \(x=-5\) and \(x = 4\)? Wait, no, the graph: when \(x=-5\), the graph touches the x - axis, and when \(x = 4\), it touches the x - axis. So the function has factors of \((x + 5)^2\) and \((x - 4)^2\)? Wait, no, wait the left touch is at \(x=-5\)? Wait, the x - axis: the left vertex is at \(x=-5\) (because it's between - 6 and - 4, so \(x=-5\)) and the right vertex is at \(x = 4\) (between 2 and 6, so \(x = 4\)). So the function can be written in the form \(f(x)=a(x + 5)^2(x - 4)^2\)? Wait, no, wait the y - intercept is at \((0,-1200)\). Let's check.

Wait, maybe the roots are \(x=-5\) and \(x = 4\) with multiplicity 2. So the general form of a quartic function (since it's a 4th - degree polynomial, as it has two "humps" and two "valleys") with double roots at \(x=-5\) and \(x = 4\) is \(f(x)=a(x + 5)^2(x - 4)^2\). Now, we can use the y - intercept to find \(a\). The y - intercept is when \(x = 0\), \(y=-1200\).

Substitute \(x = 0\) into the function:

\(f(0)=a(0 + 5)^2(0 - 4)^2=a\times25\times16=a\times400\)

We know that \(f(0)=-1200\), so:

\(400a=-1200\)

Step2: Solve for \(a\)

Divide both sides by 400: \(a=\frac{-1200}{400}=-3\)

So the function is \(f(x)=-3(x + 5)^2(x - 4)^2\). Wait, but let's expand it to check.

First, \((x + 5)^2=x^{2}+10x + 25\), \((x - 4)^2=x^{2}-8x + 16\)

Multiply \((x^{2}+10x + 25)(x^{2}-8x + 16)=x^{4}-8x^{3}+16x^{2}+10x^{3}-80x^{2}+160x + 25x^{2}-200x + 400=x^{4}+2x^{3}-39x^{2}-40x + 400\)

Then multiply by - 3: \(f(x)=-3x^{4}-6x^{3}+117x^{2}+120x - 1200\)

Wait, but let's check the y - intercept. When \(x = 0\), \(f(0)=-3\times0 - 6\times0+117\times0 + 120\times0-1200=-1200\), which matches the graph's y - intercept.

Alternatively, maybe the roots are \(x=-5\) and \(x = 4\) with multiplicity 2. Wait, another way: the graph is a quartic function (degree 4) with double roots at \(x=-5\) and \(x = 4\), and the leading coefficient negative (since the ends go downwards, as \(x ightarrow\pm\infty\), \(f(x) ightarrow-\infty\)).

Wait, maybe I made a mistake in the roots. Let's re - examine the graph. The left vertex is at \(x=-5\) (between - 6 and - 4) and the right vertex is at \(x = 4\)? Wait, no, the x - intercepts (touching) are at \(x=-5\) and \(x = 4\)? Wait, no, the left touch is at \(x=-5\) (mid - point of - 6 and - 4) and the right touch is at \(x = 4\) (mid - point of 2 and 6). So the function is a quartic with factors \((x + 5)^2\) and \((x - 4)^2\), and the leading coefficient negative (since the graph opens downward, as \(x ightarrow\pm\infty\), \(y ightarrow-\infty\)).

Wait, another approach: the general form of a quartic function with double roots at \(x = r_1\) and \(x = r_2\) is \(f(x)=a(x - r_1)^2(x - r_2)^2\). From the graph, when \(x = 0\), \(y=-1200\). Let's find the roots. The graph touches the x - axis at \(x=-5\) and \(x = 4\)? Wait, no, looking at the x - axis, the left touch is at \(x=-5\) (because the vertex is at \(x=-5\), between - 6 and - 4) and the right touch is at \(x = 4\) (vertex at \(x = 4\), between 2 and 6). So \(r_1=-5\), \(r_2 = 4\). Then \(f(x)=a(x + 5)^2(x - 4)^2\). Substitute \(x = 0\), \(y=-1200\):

\(a(5)^2(-4)^2=a\times25\times16 = 400a=-1200\), so \(a=-3\). So \(f(x)=-3(x + 5)^2(x - 4)^2\). Let's expand this:

\((x + 5)^2=x^{2}+10x + 25\), \((x - 4)^2=x^{2}-8x + 16\)

\((x^{2}+10x + 25)(x^{2}-8x + 16)=x^{4}-8x^{3}+16x^{2}+10x^{3}-80x^{2}+160x + 25x^{2}-200x + 400=x^{4}+2x^{3}-39x^{2}-40x + 400\)

Multiply by - 3: \(f(x)=-3x^{4}-6x^{3}+117x^{2}+120x - 1200\)

We can also write it in factored form as \(f(x)=-3(x + 5)^2(x - 4)^2\) or expand it. Another way: maybe the roots are \(x=-5\) and \(x = 4\), but let's check the shape. The function is a quartic, even degree, leading coefficient negative, so it opens downward, which matches the graph (both ends go down). The y - intercept is - 1200, which matches.

Answer:

\(f(x)=-3(x + 5)^2(x - 4)^2\) (or the expanded form \(-3x^{4}-6x^{3}+117x^{2}+120x - 1200\))