Find an equation of the secant line containing (1, f(1)) and (2, f(2)).
-$f ( x ) = \sqrt { x + 48 }$
A) $y = ( - 5 \sqrt { 2 } - 7 ) x - 5 \sqrt { 2 } + 14$
B) $y = ( 5 \sqrt { 2 } - 7 ) x - 5 \sqrt { 2 } + 14$
C) $y = ( - 5 \sqrt { 2 } + 7 ) x + 5 \sqrt { 2 } - 14$
D) $y = ( 5 \sqrt { 2 } - 7 ) x + 5 \sqrt { 2 } - 14$

Question

Find an equation of the secant line containing (1, f(1)) and (2, f(2)).
-$f ( x ) = \sqrt { x + 48 }$ 
A) $y = ( - 5 \sqrt { 2 } - 7 ) x - 5 \sqrt { 2 } + 14$
B) $y = ( 5 \sqrt { 2 } - 7 ) x - 5 \sqrt { 2 } + 14$
C) $y = ( - 5 \sqrt { 2 } + 7 ) x + 5 \sqrt { 2 } - 14$
D) $y = ( 5 \sqrt { 2 } - 7 ) x + 5 \sqrt { 2 } - 14$

Quizplus · Accepted Answer

First, calculate $f(1) = \sqrt{1 + 48} = \sqrt{49} = 7$ and $f(2) = \sqrt{2 + 48} = \sqrt{50} = 5\sqrt{2}$. The slope of the secant line is $\frac{5\sqrt{2} - 7}{2 - 1} = 5\sqrt{2} - 7$. Using point-slope form with point (1, 7), the equation becomes $y - 7 = (5\sqrt{2} - 7)(x - 1)$, which simplifies to $y = (5\sqrt{2} - 7)x - 5\sqrt{2} + 14$.

Find an Equation of the Secant Line Containing (1, F(1)) f(x)=x+48f ( x ) = \sqrt { x + 48 }f(x)=x+48​

Find an Equation of the Secant Line Containing (1, F(1)) $f ( x ) = \sqrt { x + 48 }$