Thread: Find the equation of the line tangent to graph f

I need to find an equation of the line that is tangent to the graph of f and parallel to the given line.

Function:
f(x)=(x^3)+2

Line:
3x-y-4=0


So far I rearranged the equation of the line to get... y=3x-4

So I plugged lines with the slope of 3 into my calculator. The graph y=3x seemed to look tangent to f(x) and parallel to the line. However, I need to solve this problem without a calculator, and without guessing by the looks of a graph.

Thanks to anyone who tries to help me.

--Cori

Originally Posted by liz155

I need to find an equation of the line that is tangent to the graph of f and parallel to the given line.

Function:
f(x)=(x^3)+2

Line:
3x-y-4=0


So far I rearranged the equation of the line to get... y=3x-4

So I plugged lines with the slope of 3 into my calculator. The graph y=3x seemed to look tangent to f(x) and parallel to the line. However, I need to solve this problem without a calculator, and without guessing by the looks of a graph.

Thanks to anyone who tries to help me.

--Cori

recall that the derivative gives the slope of the tangent line at any arbitrary point.

find and set it equal to three and solve for . that would give you the required x-coordinate. find the y-coordinate and then derive the tangent line from that

try it. if you have any questions, ask

Ok, So I set f'(x)=3x^2=3
Then I solved for x and got 1

But how do I solve for y? Do I plug the 1 back into the derivative equation?

Like this?
f'(1)=3

So the pt = (1,3) ?

Then I'm stuck on how to find the equation for the line...

Please let me know if I'm on the right track.

Thanks so much!!!

--Cori

Originally Posted by liz155

Ok, So I set f'(x)=3x^2=3
Then I solved for x and got 1

But how do I solve for y? Do I plug the 1 back into the derivative equation?

Like this?
f'(1)=3

So the pt = (1,3) ?

Then I'm stuck on how to find the equation for the line...

Please let me know if I'm on the right track.

Thanks so much!!!

--Cori

first of,

you find the y-coordinate by plug the x-value you got into the original equation. you want a point on the graph, correct? why would you plug it into the derivative equation?

when you find the desired coordinates, let's call them , then we can use them along with the slope in the point-slope form to find the equation of the line

Point-Slope Form:

and then rewrite it in the form when done

Ok, I see what I did. I don't know what I was thinking.

So I took the x= +/- 1 and plugged it into the original...

f(1)=(1)^3 +2 = 3
f(-1)=(-1)^3 +2 = 1

(1,3)
1=x_1
3=y_1

I used both the coordinates to get slope = 1

Then pt. slope form

y-3=(1)(x-1)
y=x+2 <-- final answer

Is that right??

Thanks SOOOOO much!!!

Edit: Wait nevermind, that isn't parallel to the line it needs to be parallel to. Ahh. I'm so confused now.

Originally Posted by liz155

Ok, I see what I did. I don't know what I was thinking.

So I took the x= +/- 1 and plugged it into the original...

f(1)=(1)^3 +2 = 3
f(-1)=(-1)^3 +2 = 1

(1,3)
1=x_1
3=y_1

I used both the coordinates to get slope = 1

Then pt. slope form

y-3=(1)(x-1)
y=x+2 <-- final answer

Is that right??

Thanks SOOOOO much!!!

Edit: Wait nevermind, that isn't parallel to the line it needs to be parallel to. Ahh. I'm so confused now.

what are you talking about? the slope is 3, we said that from the beginning. our objective was to find the correct x and y coordinates so that we can plug them into the point-slope form. use +1 or -1 for x. these correspond to two different lines, either will give you a solution

Problem: I plug in the 1 and -1

f(1)=3(1)-4 = -1
f(-1)=3(-1)-4 = -7

Coordinates: (1, -1) and (-1,-7)

Slope=3

But whenever I use pt slope I keep getting 3x-4, the original line equation. I don't know what to do from here.

Thanks for all your patience. I made some very silly mistakes.

--Cori

Originally Posted by liz155

Problem: I plug in the 1 and -1

f(1)=3(1)-4 = -1
f(-1)=3(-1)-4 = -7

Coordinates: (1, -1) and (-1,-7)

Slope=3

But whenever I use pt slope I keep getting 3x-4, the original line equation. I don't know what to do from here.

Thanks for all your patience. I made some very silly mistakes.

--Cori

the original curve is

if we use :



so we have

using and we have by the point-slope form:



------------> slope of one of the tangent lines with the required conditions


Now, this may be a bit trivial, but try it with



Moral: don't lose track of what equations you must use. you are mixing up the tangent line with the original equation with the conditional line that we are given

The other line is y=3x+4.

I really think I get it now. Thanks SOOO much for all your help!! I couldn't have figured it out without you!

--Cori

Originally Posted by liz155

The other line is y=3x+4.

I really think I get it now. Thanks SOOO much for all your help!! I couldn't have figured it out without you!

--Cori

yes you could.

you're welcome anyway