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

Revision as of 15:51, 18 February 2017

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

Foundations:
Equation of a tangent line
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:
$y=-1(x+2)+9$

@@ Line 7: / Line 7: @@
 |'''Equation of a tangent line'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; The equation of the tangent line to <math>f(x)</math> at the point <math>(a,b)</math> is
+|&nbsp; &nbsp; &nbsp; &nbsp; The equation of the tangent line to <math style="vertical-align: -5px">f(x)</math> at the point <math style="vertical-align: -5px">(a,b)</math> is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=m(x-a)+b</math> where <math>m=f'(a).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math> where <math style="vertical-align: -5px">m=f'(a).</math>
 |}
@@ Line 20: / Line 20: @@
 |First, we need to calculate the slope of the tangent line.
 |-
-|Let <math>f(x)=3\sqrt{-2x+5}.</math>
+|Let <math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
 |-
 |From Problem 3, we have
@@ Line 43: / Line 43: @@
 !Step 2: &nbsp;
 |-
-|Now, the tangent line has slope <math>m=-1</math>
+|Now, the tangent line has slope <math style="vertical-align: -1px">m=-1</math>
 |-
-|and passes through the point <math>(-2,9).</math>
+|and passes through the point <math style="vertical-align: -5px">(-2,9).</math>
 |-
 |Hence, the equation of the tangent line is