Question

let a and b be real numbers where a ≠ b ≠ 0. which of the following functions could represent the graph below?
$f(x)=x(x-a)^2(x-b)^4$
$f(x)=x(x-a)^3(x-b)^2$
$f(x)=x^4(x-a)(x-b)^2$
$f(x)=x^2(x-a)^5(x-b)$

Explanation:

Step1: Analyze end behavior

As $x\to+\infty$, $f(x)\to+\infty$; as $x\to-\infty$, $f(x)\to-\infty$. This means the leading term has an odd degree.

• Degree of option 1: $1+2+4=7$ (odd)
• Degree of option 2: $1+3+2=6$ (even)
• Degree of option 3: $4+1+2=7$ (odd)
• Degree of option 4: $2+5+1=8$ (even)

Eliminate options 2 and 4.

Step2: Analyze x-intercept behavior

• At the rightmost intercept ($x>0$), the graph crosses the x-axis (odd multiplicity).
• At the left two intercepts ($x<0$), the graph touches and turns around the x-axis (even multiplicity).

Step3: Match intercept multiplicities

• Option1: $x$ (multiplicity 1, odd), $(x-a)^2$ (multiplicity 2, even), $(x-b)^4$ (multiplicity 4, even). This would have one crossing intercept and two touching intercepts, matching the graph.
• Option3: $x^4$ (multiplicity 4, even), $(x-a)$ (multiplicity 1, odd), $(x-b)^2$ (multiplicity 2, even). This would have one touching intercept at $x=0$, which does not match the graph (the rightmost intercept is the crossing one).

Answer:

$f(x) = x(x - a)^2(x - b)^4$