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

Revision as of 14:51, 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a,b)} is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=m(x-a)+b} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=f'(a).}

Solution:

Step 1:
First, we need to calculate the slope of the tangent line.
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=3\sqrt{-2x+5}.}
From Problem 3, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\frac{-3}{\sqrt{-2x+5}}.}
Therefore, the slope of the tangent line is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=-1}
and passes through the point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-2,9).}
Hence, the equation of the tangent line is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=-1(x+2)+9.}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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