Difference between revisions of "009A Sample Midterm 3, Problem 4"
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| − | <span class="exam">Find the equation of the tangent line to <math style="vertical-align: -4px">y=3\sqrt{-2x+5}</math> at <math style="vertical-align: -4px">(-2,9).</math> | + | <span class="exam">Find the equation of the tangent line to <math style="vertical-align: -4px">y=3\sqrt{-2x+5}</math> at <math style="vertical-align: -4px">(-2,9).</math> |
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!Foundations: | !Foundations: | ||
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| − | | | + | |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 |
|- | |- | ||
| − | | | + | | |
|- | |- | ||
| − | | <math style="vertical-align: -5px">y=m(x-a)+b</math> where <math style="vertical-align: -5px">m=f'(a).</math> | + | | <math style="vertical-align: -5px">y=m(x-a)+b</math> where <math style="vertical-align: -5px">m=f'(a).</math> |
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|First, we need to calculate the slope of the tangent line. | |First, we need to calculate the slope of the tangent line. | ||
|- | |- | ||
| − | |Let <math style="vertical-align: -5px">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 | |From Problem 3, we have | ||
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!Step 2: | !Step 2: | ||
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| − | |Now, the tangent line has slope <math style="vertical-align: -1px">m=-1</math> | + | |Now, the tangent line has slope <math style="vertical-align: -1px">m=-1</math> |
|- | |- | ||
| − | |and passes through the point <math style="vertical-align: -5px">(-2,9).</math> | + | |and passes through the point <math style="vertical-align: -5px">(-2,9).</math> |
|- | |- | ||
|Hence, the equation of the tangent line is | |Hence, the equation of the tangent line is | ||
Revision as of 16:45, 26 February 2017
Find the equation of the tangent line to 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=3\sqrt{-2x+5}} at 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).}
| Foundations: |
|---|
| The equation of the tangent line to 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)} 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} |