which statement describes the graph of (f(x)=-4x^{3}-28x^{2}-32x + 64)?\nthe graph crosses the x - axis at…

Answer:

First, factor the polynomial (f(x)=-4x^{3}-28x^{2}-32x + 64). Factor out (-4) first: (f(x)=-4(x^{3}+7x^{2}+8x - 16)). By the Rational - Root Theorem, the possible rational roots are factors of (16), i.e., (\pm1,\pm2,\pm4,\pm8,\pm16). Let's test (x = - 4): [ \begin{align*} (-4)^{3}+7(-4)^{2}+8(-4)-16&=-64 + 112-32 - 16\ &=-64-32-16 + 112\ &=-112 + 112\ &=0 \end{align*} ] Since (x=-4) is a root, ((x + 4)) is a factor of (x^{3}+7x^{2}+8x - 16). Using polynomial long - division or synthetic division: Dividing (x^{3}+7x^{2}+8x - 16) by ((x + 4)) gives (x^{2}+3x - 4). Factor (x^{2}+3x - 4=(x + 4)(x - 1)). So, (f(x)=-4(x + 4)^{2}(x - 1)).

The roots of the function are (x=-4) and (x = 1). If a factor ((x - a)) has an odd multiplicity, the graph of the function crosses the (x) - axis at (x=a). If a factor ((x - a)) has an even multiplicity, the graph of the function touches the (x) - axis at (x=a). The factor ((x + 4)) has a multiplicity of (2) (even), and the factor ((x - 1)) has a multiplicity of (1) (odd). So the graph touches the (x) - axis at (x=-4) and crosses the (x) - axis at (x = 1). The correct option is: The graph touches the (x) - axis at (x=-4) and crosses the (x) - axis at (x = 1).

Explanation:

Step1: Factor out common factor

(f(x)=-4(x^{3}+7x^{2}+8x - 16))

Step2: Find a root using Rational - Root Theorem

Test (x=-4): ((-4)^{3}+7(-4)^{2}+8(-4)-16 = 0)

Step3: Divide the polynomial

Divide (x^{3}+7x^{2}+8x - 16) by ((x + 4)) to get (x^{2}+3x - 4)

Step4: Factor the quadratic

(x^{2}+3x - 4=(x + 4)(x - 1))

Step5: Determine multiplicity and graph behavior

(f(x)=-4(x + 4)^{2}(x - 1)), ((x + 4)) has even multiplicity, ((x - 1)) has odd multiplicity.