Question

the function f(x)=1.1x³ - 33x² + 260x + 564 below models the number of discharges from the military, f(x), of active - duty gay service members under the \dont ask, dont tell\ policy x years after 1994. complete parts a and b. a. find the slope of the secant line from x₁ = 0 to x₂ = 4. the slope of the secant line is . (round to the nearest whole number as needed.)

Explanation:

Step1: Calculate f(0)

Substitute \(x = 0\) into \(f(x)=1.1x^{3}-33x^{2}+260x + 564\).

$$f(0)=1.1\times0^{3}-33\times0^{2}+260\times0 + 564=564$$

Step2: Calculate f(4)

Substitute \(x = 4\) into \(f(x)=1.1x^{3}-33x^{2}+260x + 564\).

$$
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Step3: Calculate the slope of the secant line

The slope \(m\) of the secant line between \((x_1,f(x_1))\) and \((x_2,f(x_2))\) is given by \(m=\frac{f(x_2)-f(x_1)}{x_2 - x_1}\). Here \(x_1 = 0\), \(x_2=4\), \(f(x_1)=564\), \(f(x_2)=1146.4\).

$$m=\frac{f(4)-f(0)}{4 - 0}=\frac{1146.4 - 564}{4}=\frac{582.4}{4}=145.6$$

Rounding to the nearest whole - number, we get \(m = 146\).

Answer:

146