given $f(x)=x^{3}-5$, find the equation of the secant line passing through $(-3,f(-3))$ and $(2,f(2))$…

given $f(x)=x^{3}-5$, find the equation of the secant line passing through $(-3,f(-3))$ and $(2,f(2))$. write your answer in the form $y = mx + b$.

given $f(x)=x^{3}-5$, find the equation of the secant line passing through $(-3,f(-3))$ and $(2,f(2))$. write your answer in the form $y = mx + b$.

Answer

Explanation:

Step1: Calculate $f(-3)$ and $f(2)$

First, find $f(-3)$: [ \begin{align*} f(-3)&=(-3)^3 - 5\ &=-27 - 5\ &=-32 \end{align*} ] Then, find $f(2)$: [ \begin{align*} f(2)&=2^3 - 5\ &=8 - 5\ &=3 \end{align*} ] So the two points are $(-3,-32)$ and $(2,3)$.

Step2: Calculate the slope $m$

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Here, $x_1=-3,y_1 = - 32,x_2=2,y_2 = 3$. [ \begin{align*} m&=\frac{3-(-32)}{2-(-3)}\ &=\frac{3 + 32}{2+3}\ &=\frac{35}{5}\ &=7 \end{align*} ]

Step3: Find the y - intercept $b$

Use the point - slope form $y - y_1=m(x - x_1)$ with the point $(2,3)$ and $m = 7$. [ \begin{align*} y-3&=7(x - 2)\ y-3&=7x-14\ y&=7x-14 + 3\ y&=7x-11 \end{align*} ]

Answer:

$y = 7x-11$