Question

let a and b be real numbers where a ≠ b ≠ 0. which of the following functions could represent the graph below? graph of a function with two minima on the x - axis (not at the origin) and a maximum on the y - axis, opening upwards options: - $f(x)=x(x - a)^3(x - b)^3$ - $f(x)=(x - a)^2(x - b)^4$ - $f(x)=x(x - a)^6(x - b)^2$ - $f(x)=(x - a)^5(x - b)$

Explanation:

Step1: Analyze x-intercepts and multiplicity

The graph touches the x - axis at two points (the roots with even multiplicity) and has a y - intercept (when \(x = 0\)) that is positive (since the graph is above the x - axis at \(x=0\)). Let's analyze each option:

Option 1: \(f(x)=x(x - a)^{3}(x - b)^{3}\)

• The root at \(x = 0\) has multiplicity 1 (odd), so the graph should cross the x - axis at \(x = 0\). But the given graph does not cross the x - axis at \(x = 0\) (it is above the x - axis at \(x = 0\)), so this option is incorrect.

Option 2: \(f(x)=(x - a)^{2}(x - b)^{4}\)

• The roots are \(x=a\) (multiplicity 2, even) and \(x = b\) (multiplicity 4, even). So the graph will touch the x - axis at \(x=a\) and \(x = b\) (since even multiplicity means the graph touches the axis and turns around). Also, when \(x = 0\), \(f(0)=( - a)^{2}( - b)^{4}=a^{2}b^{4}\). Since \(a

eq0\) and \(b
eq0\), \(a^{2}>0\) and \(b^{4}>0\), so \(f(0)>0\), which matches the y - intercept (the graph is above the x - axis at \(x = 0\)). Also, the degree of the polynomial is \(2 + 4=6\) (even), and the leading coefficient (when expanded, the leading term will have a positive coefficient since the leading terms of \((x - a)^{2}\) is \(x^{2}\) and of \((x - b)^{4}\) is \(x^{4}\), so the leading term is \(x^{2}\times x^{4}=x^{6}\) with coefficient 1 (positive)). For even degree with positive leading coefficient, as \(x
ightarrow\pm\infty\), \(f(x)
ightarrow+\infty\), which matches the end - behavior of the graph (both ends go up).

Option 3: \(f(x)=x(x - a)^{6}(x - b)^{2}\)

• The root at \(x = 0\) has multiplicity 1 (odd), so the graph should cross the x - axis at \(x = 0\), but the given graph does not cross the x - axis at \(x = 0\), so this option is incorrect.

Option 4: \(f(x)=(x - a)^{5}(x - b)\)

• The root \(x=a\) has multiplicity 5 (odd) and \(x = b\) has multiplicity 1 (odd). So the graph will cross the x - axis at \(x=a\) and \(x = b\) (since odd multiplicity means the graph crosses the axis), but the given graph touches the x - axis at the roots (does not cross), so this option is incorrect.

Answer:

\(f(x)=(x - a)^{2}(x - b)^{4}\)