Thread: Solving Equation of Tangent Line to Curve

I'm having trouble solving this problem. I have to find the equation of a line tangent to a curve at a specific point. I also have to use a certain technique.

The curve is:

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

The point to be tangent to is:
P = (-1, 6)

I have to use these equations to solve.

m(tan) = lim(x->c) (f(x) - f(c))/(x-c))

and

y - f(c) = m(tan)(x - c)

I can get to the point where I have (x^2 - 2x) / (x+1). I'm not sure how to proceed. Thanks

Originally Posted by salted

I'm having trouble solving this problem. I have to find the equation of a line tangent to a curve at a specific point. I also have to use a certain technique.

The curve is:

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

The point to be tangent to is:
P = (-1, 6)

As I understand this question you are asked to find the equation
of the tangent to at the point .

The problem with this is that is not on the curve.

RonL

Good point. Perhaps the OP meant the point (1,6)?

Originally Posted by salted

I'm having trouble solving this problem. I have to find the equation of a line tangent to a curve at a specific point. I also have to use a certain technique.

The curve is:

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

The point to be tangent to is:
P = (-1, 6)

I have to use these equations to solve.

m(tan) = lim(x->c) (f(x) - f(c))/(x-c))

and

y - f(c) = m(tan)(x - c)

I can get to the point where I have (x^2 - 2x) / (x+1). I'm not sure how to proceed. Thanks

OK lets find the tangent at

,

,

so:



hence:



Which is the horizontal line

-anyway this shows the technique of finding the slope you are required to
use.

RonL

OK now lets find the tangent at

,

,

so:



hence:



So the tangent at has slope , and
applying your second equation gives the equation of the line as:



where and so:

,

or:



RonL