# Find equation of line f(x) parallel to given line

• Jul 30th 2009, 03:48 PM
drkidd22
Find equation of line f(x) parallel to given line
This is another one of those problems that I kinda get the idea of how to do it but then just can't think of anything else. So, to the problem.

Function
$
f(x) = x^3
$

$f'(x) = 3x^2$

Line
$
3x-y+1 = 0
$

$y = 3x+1$

From here we know that $m=3$ since $y=mx+b$

Then I made $f'(x)=3$
$3x^2=3$
$x^2=1$

To get the value of x.
$x=1$

Now I go back to the given line and plug in:
$y=3(1)+1$
$y=4$

My point is $(1,4)$

I did most of it as I was writing this, I think my light bulb turned on.

$y-y1=m(x-x1)$
$y-4=3(x-1)$
$y=3x+1$

Did I do it right or is there another way of doing this?
• Jul 30th 2009, 03:58 PM
skeeter
Quote:

Originally Posted by drkidd22
This is another one of those problems that I kinda get the idea of how to do it but then just can't think of anything else. So, to the problem.

Function
$
f(x) = x^3
$

$f'(x) = 3x^2$

Line
$
3x-y+1
$
is this 3x-y+1 = 0 ?
$y = 3x+1$

From here we know that $m=3$ since $y=mx+b$

Then I made $f'(x)=3$
$3x^2=3$
$x^2=1$

To get the value of x.
$x=1$

x = -1 also

Now I go back to the given line and plug in:
$y=3(1)+1$ 3(1) + 1 = 4, not 5
$y=5$

.
• Jul 30th 2009, 04:03 PM
Haytham
Quote:

Originally Posted by drkidd22
This is another one of those problems that I kinda get the idea of how to do it but then just can't think of anything else. So, to the problem.

Function
$
f(x) = x^3
$

$f'(x) = 3x^2$

Line
$
3x-y+1
$

$y = 3x+1$

From here we know that $m=3$ since $y=mx+b$

Then I made $f'(x)=3$
$3x^2=3$
$x^2=1$

To get the value of x.
$x=1$

Now I go back to the given line and plug in:
$y=3(1)+1$
y=5

My point is $(1,5)$

I did most of it as I was writing this, I think my light bulb turned on.

$y-y1=m(x-x1)$
$y-5=3(x-1)$
$y=3x+2$

Did I do it right or is there another way of doing this?

red part is wrong
if I understood the question you should enter the $x=1$ in the $
f(x) = x^3
$
function, instead of the line equation. the line equation will give you the line equation, if you correct the red part
• Jul 30th 2009, 04:04 PM
drkidd22
• Jul 30th 2009, 04:08 PM
drkidd22
http://www.mathhelpforum.com/math-he...8f41dd91-1.gif this is the derrivative of $f(x)=x^3$ which you need to find the slope, since the slope is 3 that why I made http://www.mathhelpforum.com/math-he...8f41dd91-1.gif=3 to get x = 1
• Jul 30th 2009, 04:27 PM
skeeter

if
$x^2 = 1$ , then $x^2 - 1 = 0$ ... $(x+1)(x-1) = 0$

this says that there will be two distinct lines tangent to the curve y = x^3 that are parallel to
$y = 3x+1$

• Jul 30th 2009, 05:13 PM
drkidd22
I'm not complettly clear on this problem yet. So what is the equation of the line parallel to the given line? and how is it done?
• Jul 30th 2009, 05:26 PM
skeeter
Quote:

Originally Posted by drkidd22
I'm not complettly clear on this problem yet. So what is the equation of the line parallel to the given line? and how is it done?

you've established that the slope of a line parallel to y = 3x+1 is 3.

$f'(x) = 3$

$3x^2 = 3$

$x = \pm 1$

now ... get the two points on the original curve where f'(x) = 3

$y = x^3$

$y = (1)^3$ ... (1,1)

y - 1 = 3(x - 1)

y = 3x - 2 is the first line

$y = (-1)^3$ ... (-1,-1)

y + 1 = 3(x + 1)

y = 3x + 2 is the second line

now look at the attached graph ... (I made the colors match)
• Jul 30th 2009, 05:45 PM
drkidd22
Thanks alot, that clears it up. May I ask what program/software you used to make the graph?
• Jul 31st 2009, 05:39 AM
skeeter
Quote:

Originally Posted by drkidd22
Thanks alot, that clears it up. May I ask what program/software you used to make the graph?

Graph