Hi,
If I was told to find the equation of line with slope (1) that is tangent to the curve of a rational function, 1/-x-3 , would I get two tangent lines at the end?
Important lease do not use calculus or derivatives.
Here are the steps I use...
I first find the general equation of the line, which would be y=(1)x+b
then i let the rational function = the linear function
Then I rearrange the equation so that everything is on the left side and zero is on the left side.
Then, I make the equation be in the standard form of a quadratic,
Then I use the discriminant and let it equal zero, (previous notes that i have say :so that we may have one solution to the quadratic,one double root)
Then solve for b
However at the end My answer looks like this:
(b-5)(b-1)=0
so b=5 or b=1
so that way i would have two equations:
y=x+5 and y=x+1
Although it makes perfect sense that they are tangents ,however, doesnt this mean that i have two solutions ?
Thank you
If I was told to find the equation of line with slope (1) that is tangent to the curve of a rational function, 1/-x-3 , would I get two tangent lines at the end?
Important lease do not use calculus or derivatives.
Here are the steps I use...
I first find the general equation of the line, which would be y=(1)x+b
then i let the rational function = the linear function
Then I rearrange the equation so that everything is on the left side and zero is on the left side.
Then, I make the equation be in the standard form of a quadratic,
Then I use the discriminant and let it equal zero, (previous notes that i have say :so that we may have one solution to the quadratic,one double root)
Then solve for b
However at the end My answer looks like this:
(b-5)(b-1)=0
so b=5 or b=1
so that way i would have two equations:
y=x+5 and y=x+1
Although it makes perfect sense that they are tangents ,however, doesnt this mean that i have two solutions ?
Thank you