Difference between revisions of "009A Sample Midterm 3, Problem 4"

Revision as of 16:26, 17 February 2017

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

Foundations:
Tangent line formula

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

@@ Line 14: / Line 14: @@
 !Step 1: &nbsp;
 |-
-|
+|First, we need to calculate the slope of the tangent line.
+|-
+|Let <math>f(x)=3\sqrt{-2x+5}.</math>
+|-
+|From Problem 3, we have
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=\frac{-3}{\sqrt{-2x+5}}.</math>
+|-
+|Therefore, the slope of the tangent line is
 |-
 |
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
+\displaystyle{m} & = & \displaystyle{f'(-2)}\\
+&&\\
+& = & \displaystyle{\frac{-3}{\sqrt{-2(-2)+5}}}\\
+&&\\
+& = & \displaystyle{\frac{-3}{\sqrt{9}}}\\
+&&\\
+& = & \displaystyle{-1.}
+\end{array}</math>
 |}
@@ Line 22: / Line 39: @@
 !Step 2: &nbsp;
 |-
-|
+|Now, the tangent line has slope <math>m=-1</math>
+|-
+|and passes through the point <math>(-2,9).</math>
+|-
+|Hence, the equation of the tangent line is
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=-1(x+2)+9.</math>
 |}
@@ Line 31: / Line 52: @@
 !Final Answer: &nbsp;
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=-1(x+2)+9</math>
 |-
 |
 |}
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]