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Find line tangent to curve which is parallel to other line

  • #1

Homework Statement


Find the line tangent to the curve f(x)=0.5x2+3x-1 which is parallel to the line g(x)=x/2+0.5

Homework Equations


f'(x)=x+3

The Attempt at a Solution


I know it involves taking the derivative of f(x) and using it somehow, but I don't know where to go from there.
 

Answers and Replies

  • #2
Math_QED
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Homework Statement


Find the line tangent to the curve f(x)=0.5x2+3x-1 which is parallel to the line g(x)=x/2+0.5

Homework Equations


f'(x)=x+3

The Attempt at a Solution


I know it involves taking the derivative of f(x) and using it somehow, but I don't know where to go from there.
You want to find the tangent line to the curve ##f(x)##, parallel to the line ##g(x)##. If the tangent line has to be parallel with ##g(x)##, what do you know about the slope of this line?
 
  • #3
Both lines will need to have the same slope, 1/2.
 
  • #4
Math_QED
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Both lines will need to have the same slope, 1/2.
Correct. Now, the derivative gives the slope of the tangent line in each point. Where is the slope ##1/2##?
Can you find the equation of the tangent line then?
 
  • #5
The slope is 1/2 when f'(x)=1/2:
1/2=x+3
x=-5/2
 
  • #6
Math_QED
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The slope is 1/2 when f'(x)=1/2:
1/2=x+3
x=5/2
Minor mistake in your arithmetic.

You have a slope and a point (if you fix the mistake). There is a unique line that fits those 2 conditions. Can you find the equation?
 
  • #7
Yes caught the mistake.

So using -5/2 on f(x) gets f(-5/2)=-43/8, which means a point on the line is (-5/2, -43/8).
If the line is of the form y=1/2*x+b, we substitute the point for x and y:
-43/8=1/2*-5/2+b
b=-33/8
The line is: y=1/2*x-33/8
 
  • #8
Math_QED
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Yes caught the mistake.

So using -5/2 on f(x) gets f(-5/2)=-43/8, which means a point on the line is (-5/2, -43/8).
If the line is of the form y=1/2*x+b, we substitute the point for x and y:
-43/8=1/2*-5/2+b
b=-33/8
The line is: y=1/2*x-33/8
That's correct! Well done!
 
  • #9
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Thread moved. @bubblescript, please post problems that involve derivatives in the Calculus & Beyond section.
 

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