line that is tangent to a curve ata particular point

Find the line that is tangent to the curve with the function y^2=x+3 at point (6,-3)

I know m=dy/dx so implicit derivative is: 2y dy/dx=1, so dy/dx = 1/2y

then equation of a line y-y_1=m(x-x_1). thus y+3=1/2y(x-6)

I can't believe that this is suppose to be my answer as this is clearly not a straight line. What am I doing wrong?

Thanks

Quote:

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Originally Posted by Bowlbase

Find the line that is tangent to the curve with the function y^2=x+3 at point (6,-3)
I know m=dy/dx so implicit derivative is: 2y dy/dx=1, so dy/dx = 1/2y
then equation of a line y-y_1=m(x-x_1). thus y+3=1/2y(x-6)
I can't believe that this is suppose to be my answer as this is clearly not a straight line. What am I doing wrong?

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At the point we have .

Really, that simple? Geez..

Thanks

Rather than make a new thread I have similar question as before but no points given

Find all the lines tangent to the curve y=x^3 and is parallel to y=3x-5.

I have that the slope must be the same as the given line, so m=3. The derivative of the curve also gives slope so 3x^2=3 where x is +-1.

So I have only 1 point to find in my line equation. I have y-y_1=3(x+-1). I'm stuck on trying to figure out what y_1 is here.

Pretty sure this is going to be another oh geez moment. Thanks for the help.

Quote:

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Originally Posted by Bowlbase

Rather than make a new thread I have similar question as before but no points given
Find all the lines tangent to the curve y=x^3 and is parallel to y=3x-5.
I have that the slope must be the same as the given line, so m=3. The derivative of the curve also gives slope so 3x^2=3 where x is +-1.
So I have only 1 point to find in my line equation. I have y-y_1=3(x+-1). I'm stuck on trying to figure out what y_1 is here.

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Always, always start a new thread for a new question!

You have two points of tangency
So you have two tangent lines.

Sorry, I'll make a new thread next time.

So I understand that there will be two lines but I don't see where the y values are coming from. I am assuming that you just put the x values into the y=x^3. Is this correct?

Quote:

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Originally Posted by Bowlbase

So I understand that there will be two lines but I don't see where the y values are coming from. I am assuming that you just put the x values into the y=x^3. Is this correct?

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since the tangent line "touches" the curve at the point of tangency, that would be correct.

makes perfect sense, thank you.