MathAdmin at 02:26, 14 April 2015

2015-04-14T02:26:55Z

MathAdmin: Created page with "Find the slope of the tangent line to the graph of $f(x)=x^{3}-3x^{2}-5x+7$ at the point
2015-04-14T02:25:16Z

Created page with "<span class="exam">Find the slope of the tangent line to the graph of <math style="vertical-align: -14%">f(x)=x^{3}-3x^{2}-5x+7</math> at the point <math style="vertical-align..."

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<span class="exam">Find the slope of the tangent line to the graph of <math style="vertical-align: -14%">f(x)=x^{3}-3x^{2}-5x+7</math>
at the point <math style="vertical-align: -14%">(3,-8)</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
|-
|Recall that for a given value, <math style="vertical-align: -16%">f'(x)</math> is precisely the point of the tangent line through the point <math style="vertical-align: -16%">\left(x,f(x)\right)</math>. Once we have the slope, we can then use the point-slope form for a line:
|-
|
::<math>y-y_0 = m(x-x_0),</math>
|-
|where <math style="vertical-align: 0%">m</math> is the known slope and <math style="vertical-align: -16%">\left(x_0,y_0\right)</math> is a point on the line.
|}

 '''Solution:'''

{|class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Finding the slope:  
|-
|Note that
|-
|
::<math>f'(x)\,\,=\,\,3x^2-6x-5,</math>
|-
|so the tangent line through <math style="vertical-align: -20%">(3,-8)</math> has slope
|-
|
::<math>m\,\,=\,\,f'(3)\,\,=\,\,3(3)^2-6(3)-5\,\,=\,\,4.</math>
|}

{|class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Write the Equation of the Line:  
|-
|Using the point-slope form listed in foundations, along with the point <math style="vertical-align: -20%">(3,-8)</math> and the slope <math style="vertical-align: 0%">m=4</math>, we find
|-
|
::<math>y-(-8)\,\,=\,\,4(x-3),</math>
|-
|or
|-
|
::<math>y\,\,=\,\,4x-20.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
::<math>y\,\,=\,\,4x-20.</math>
|}

[[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 5: / Line 5: @@
 ! Foundations: &nbsp;
 |-
-|Recall that for a given value, <math style="vertical-align: -16%">f'(x)</math> is precisely the point of the tangent line through the point <math style="vertical-align: -16%">\left(x,f(x)\right)</math>. Once we have the slope, we can then use the point-slope form for a line:
+|Recall that for a given value, <math style="vertical-align: -18%">f'(x)</math> is precisely the point of the tangent line through the point <math style="vertical-align: -16%">\left(x,f(x)\right)</math>. Once we have the slope, we can then use the point-slope form for a line:
 |-
+|
@@ Line 32: / Line 32: @@
 !Write the Equation of the Line: &nbsp;
 |-
-|Using the point-slope form listed in foundations, along with the point <math style="vertical-align: -20%">(3,-8)</math> and the slope <math style="vertical-align: 0%">m=4</math>, we find
+|Using the point-slope form listed in foundations, along with the point <math style="vertical-align: -20%">(3,-8)</math> and the slope <math style="vertical-align: -3%">m=4</math>, we find
 |-
+|