Question

let a and b be real numbers, where a ≠ b ≠ 0. which of the following functions could represent the graph?
$f(x) = (x - a)(x - b)^2$
$f(x) = x(x - a)(x - b)^2$
$f(x) = x^2(x - a)^2(x - b)^2$
$f(x) = x(x - a)^3(x - b)$

Explanation:

Step1: Analyze End Behavior

The leading term's degree and coefficient determine end behavior. For a polynomial, if the degree is odd, ends go opposite; if even, same. The graph has both ends up (even degree? Wait, no—wait, the right end: as $x\to\infty$, the graph goes up, left end also up. Wait, no, wait the left end: when $x\to -\infty$, the graph goes up, right end $x\to\infty$ also up. Wait, no, looking at the graph: left side, as $x$ decreases (goes left), the graph goes up; right side, as $x$ increases, goes up. So leading coefficient positive, even degree? Wait no, wait the options: let's check degrees.

Option 1: $f(x)=(x - a)(x - b)^2$. Degree: $1 + 2 = 3$ (odd). End behavior: as $x\to\infty$, $f(x)\to\infty$; $x\to -\infty$, $f(x)\to -\infty$. But graph has both ends up, so this is out.

Option 2: $f(x)=x(x - a)(x - b)^2$. Degree: $1 + 1 + 2 = 4$ (even). Leading coefficient: $1\times1\times1 = 1$ (positive). So end behavior: $x\to\pm\infty$, $f(x)\to\infty$ (matches graph's both ends up).

Option 3: $f(x)=x^2(x - a)^2(x - b)^2$. Degree: $2 + 2 + 2 = 6$ (even), leading coefficient positive. But let's check roots. This polynomial has roots at $x=0$, $x=a$, $x=b$, all with even multiplicity (so graph touches x-axis at roots, no crossing). But the given graph crosses the x-axis (at least one root with odd multiplicity). So this is out (all roots even multiplicity, graph touches, not crosses).

Option 4: $f(x)=x(x - a)^3(x - b)$. Degree: $1 + 3 + 1 = 5$ (odd). End behavior: $x\to\infty$, $f(x)\to\infty$; $x\to -\infty$, $f(x)\to -\infty$ (opposite ends), which doesn't match graph (both ends up). So out.

Step2: Analyze Root Multiplicities

Roots: where graph crosses or touches x-axis. The graph crosses the x-axis (so odd multiplicity) and touches (even multiplicity)? Wait, the graph: let's see the roots. There's a root at $x=2$ (crosses? Wait, no, at $x=2$, the graph touches? Wait, no, the graph at $x=2$: it comes down, touches? Wait, no, the right root: the graph has a root at $x=2$ (the rightmost root), and another root? Wait, the left side: a root where it touches (even multiplicity) and then crosses? Wait, no, let's look at the graph. The graph has a root with even multiplicity (touches, turns around) and a root with odd multiplicity (crosses). Wait, the option 2: $f(x)=x(x - a)(x - b)^2$. Let's say $x=0$ (root from $x$), $x=a$ (root from $(x - a)$), $x=b$ (root from $(x - b)^2$). So $x=b$ has multiplicity 2 (even, touches x-axis), $x=0$ and $x=a$ have multiplicity 1 (odd, cross x-axis). Which matches: one root with even multiplicity (touches) and two with odd (cross). The graph shows a root that touches (turns around) and others that cross. So option 2 fits.

So the correct function is $f(x)=x(x - a)(x - b)^2$.

Answer:

$f(x) = x(x - a)(x - b)^2$