009A Sample Midterm 3, Problem 4

Find the equation of the tangent line to  $y=3{\sqrt {-2x+5}}$  at  $(-2,9).$

Foundations:
The equation of the tangent line to  $f(x)$  at the point  $(a,b)$  is
$y=m(x-a)+b$  where  $m=f'(a).$

Solution:

Step 1:
First, we need to calculate the slope of the tangent line.
Let  $f(x)=3{\sqrt {-2x+5}}.$
From Problem 3, we have
$f'(x)=-{\frac {3}{\sqrt {-2x+5}}}.$
Therefore, the slope of the tangent line is
${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {f'(-2)}\\&&\\&=&\displaystyle {-{\frac {3}{\sqrt {-2(-2)+5}}}}\\&&\\&=&\displaystyle {-{\frac {3}{\sqrt {9}}}}\\&&\\&=&\displaystyle {-1.}\end{array}}$

Step 2:
Now, the tangent line has slope  $m=-1$
and passes through the point  $(-2,9).$
Hence, the equation of the tangent line is
$y=-1(x+2)+9.$

Final Answer:
$y=-1(x+2)+9$

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