Question

which of the following could be the equation of the polynomial graph shown below? (1) y=-x(x + 5)(x - 3) (2) y=x(x + 3)(x - 5) (3) y=-x²(x + 5)(x - 3) (4) y=x²(x + 3)(x - 5)

Explanation:

Step1: Identify x - intercepts

The x - intercepts of a polynomial function \(y = f(x)\) are the values of \(x\) for which \(y = 0\). For a polynomial in factored form \(y=a(x - r_1)(x - r_2)\cdots(x - r_n)\), the x - intercepts are \(x = r_1,r_2,\cdots,r_n\). The graph has x - intercepts at \(x=-5,x = 0,x = 3\).

Step2: Analyze the multiplicity of the root at \(x = 0\)

Since the graph touches the x - axis at \(x = 0\) and then turns back, the multiplicity of the root \(x = 0\) is even.

Step3: Analyze the sign of the leading - coefficient

As \(x\to+\infty\), \(y\to-\infty\), so the leading - coefficient of the polynomial is negative.

Step4: Check each option

• Option (1): \(y=-x(x + 5)(x - 3)\) has roots \(x=-5,x = 0,x = 3\) and a negative leading - coefficient. But the multiplicity of \(x = 0\) is 1 (odd).
• Option (2): \(y=x(x + 3)(x - 5)\) has roots \(x=-3,x = 0,x = 5\), which does not match the x - intercepts of the given graph.
• Option (3): \(y=-x^{2}(x + 5)(x - 3)\) has roots \(x=-5,x = 0,x = 3\), the multiplicity of \(x = 0\) is 2 (even), and the leading - coefficient is negative.
• Option (4): \(y=x^{2}(x + 3)(x - 5)\) has roots \(x=-3,x = 0,x = 5\) and a positive leading - coefficient.

Answer:

(3) \(y=-x^{2}(x + 5)(x - 3)\)