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?
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.
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?