# Solving Equation of Tangent Line to Curve

• Feb 4th 2006, 04:45 PM
salted
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
• Feb 5th 2006, 04:52 AM
CaptainBlack
Quote:

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 $f(x)=x^2+2x+3$ at the point $P=(-1, 6)$.

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

RonL
• Feb 5th 2006, 07:13 AM
Jameson
Good point. Perhaps the OP meant the point (1,6)?
• Feb 5th 2006, 11:09 AM
CaptainBlack
Quote:

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

$x=-1$,

$f(-1)=1-2+3=2$,

so:

$\frac{f(x)-f(-1)}{x-(-1))}=\frac{x^2+2x+1}{x+1}=x+1$

hence:

$\lim_{x\rightarrow -1}\frac{f(x)-f(-1)}{x-(-1))}=\lim_{x\rightarrow -1}(x+1)=0$

Which is the horizontal line $y=2$

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

RonL
• Feb 5th 2006, 12:45 PM
CaptainBlack
OK now lets find the tangent at

$x=1$,

$f(1)=1+2+3=6$,

so:

$\frac{f(x)-f(1)}{x-1}=\frac{x^2+2x-3}{x-1}=x+3$

hence:

$\lim_{x\rightarrow -1}\frac{f(x)-f(1)}{x-1}=\lim_{x\rightarrow 1}(x+3)=4$

So the tangent at $P=(1,6)$ has slope $4$, and
applying your second equation gives the equation of the line as:

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

where $c=1$ and $f(c)=6$ so:

$y-6=4(x-1)$,

or:

$y=4x+2$

RonL