Find an equation of the secant line containing (1, f(1)) and (2, f(2)).
-$$f ( x ) = \frac { 4 } { x + 3 }$$
A) $y = \frac { 1 } { 5 } x + \frac { 4 } { 5 }$

B) $y = \frac { 4 } { 5 } x + \frac { 1 } { 5 }$

C) $y = - \frac { 1 } { 5 } x + \frac { 6 } { 5 }$

D) $y = \frac { 1 } { 5 } x + \frac { 3 } { 2 }$

Question

Find an equation of the secant line containing (1, f(1)) and (2, f(2)).
-$$f ( x ) = \frac { 4 } { x + 3 }$$
A) $y = \frac { 1 } { 5 } x + \frac { 4 } { 5 }$

B) $y = \frac { 4 } { 5 } x + \frac { 1 } { 5 }$

C) $y = - \frac { 1 } { 5 } x + \frac { 6 } { 5 }$

D) $y = \frac { 1 } { 5 } x + \frac { 3 } { 2 }$

Quizplus · Accepted Answer

The slope of the secant line is calculated as $\frac{f(2) - f(1)}{2 - 1}$. Substituting $f(x) = \frac{4}{x+3}$, we find $f(1) = \frac{4}{4} = 1$ and $f(2) = \frac{4}{5}$. The slope is $\frac{\frac{4}{5} - 1}{1} = -\frac{1}{5}$. Using point-slope form with point (1, 1), the equation is $y - 1 = -\frac{1}{5}(x - 1)$, simplifying to $y = -\frac{1}{5}x + \frac{6}{5}$.

Find an Equation of the Secant Line Containing (1, F(1)) f(x)=4x+3f ( x ) = \frac { 4 } { x + 3 }f(x)=x+34​

Find an Equation of the Secant Line Containing (1, F(1)) $f ( x ) = \frac { 4 } { x + 3 }$