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

1. ## 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?

2. 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
$y=5$

.

3. 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

4. is this 3x-y+1 = 0 ? yes

To get the value of x.

x = -1 also why is this?

3(1) + 1 = 4, not 5 My bad, but then if you say x = -1 it would be 3(-1)+1= -2?

5. 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 =3 to get x = 1

6. Originally Posted by drkidd22
is this 3x-y+1 = 0 ? yes

To get the value of x.

x = -1 also why is this?

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
$y = 3x+1$

7. 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?

8. 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)

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

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