# Find line tangent to curve which is parallel to other line

## 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

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

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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

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?

Both lines will need to have the same slope, 1/2.

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?

The slope is 1/2 when f'(x)=1/2:
1/2=x+3
x=-5/2

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?

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

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!

Mark44
Mentor
Thread moved. @bubblescript, please post problems that involve derivatives in the Calculus & Beyond section.