# Algebraically, find the equation of the tangent at x=-2 for the equation, f(X)=`sqrt(x+4)`

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**Answer:** Slope of the tangent line is `sqrt(2)/4~~0.35`

If you graph this, you will notice that it is a side-opening parabola:

I've also plotted the secant line that connects (-4,0) and (0,2). Notice that these two points are two units away from our target point x=-2.

Algebraically, we take the limit as the distance between the two points we've selected approach zero. Meaning that we keep taking secant lines centered around x=-2, closer and closer together until we end up with the tangent line at x=-2. We summarize this as:

Let

`y=f(x)=sqrt(x+4)`

and ∆y is how far away a new value would be if we added a small ∆x

`y+Deltay=f(x+Deltax)rArr Deltay = f(x+Deltax)-y`

so

The slope of the secant line is defined as

`(Deltay)/(Deltax)`

The slope of the tangent line at x is when ∆x approaches zero. This is known as taking the limit.

`lim_(Deltax->0)((Deltay)/(Deltax))rArrlim_(Deltax->0)((sqrt(x+Deltax+4)-sqrt(x+4))/(Deltax))`

Now, the tricky part. There are three ways to solve limits: substitution, factoring, and the conjugate method. For this one, we need the conjugate method. We want to multiply by "1" so we don't change the fraction:

`lim_(Deltax->0)((sqrt(x+Deltax+4)-sqrt(x+4))/(Deltax)times((sqrt(x+Deltax+4)+sqrt(x+4))/(sqrt(x+Deltax+4)+sqrt(x+4))))`

It looks nasty at first, but stay with me! When you multiply conjugates, you get a difference of perfect SQUARES, which will make short work of the square roots on top... leaving us with a way out!

`lim_(Deltax->0)((x+Deltax+4-(x+4))/(Deltax(sqrt(x+Deltax+4)+sqrt(x+4))))`

Simplifying that down to:

`lim_(Deltax->0)((Deltax)/(Deltax(sqrt(x+Deltax+4)+sqrt(x+4))))`

Cancelling the ∆x on top and bottom, and applying the limit, we are left with:

`lim_(Deltax->0)((1)/(sqrt(x+Deltax+4)+sqrt(x+4)))rArr(1)/(2sqrt(x+4))`

I think it's fantastic that such a messy problem leads to a simpler solution.

Now, you must evaluate at x=-2

`(1)/(2sqrt((-2)+4))=sqrt(2)/4~~0.35`

The equation of the tangent to `f(x) = sqrt(x + 4)` at x = -2 has to be determined.

The value of the derivative of f(x) at x = -2 gives the slope of the tangent. `f'(x) = 1/(2*sqrt(x+4))` . `f'(-2) = 1/(2*sqrt 2)`

For x = -2. y = `sqrt(2)`

The equation of the tangent is `(y - sqrt 2)/(x + 2) = 1/(2*sqrt 2)`

=> `2*sqrt 2(y - sqrt 2) = x + 2`

=> `2*sqrt 2*y - 4 = x + 2`

=> `x - 2*sqrt 2*y + 6 = 0`

**The required equation of the tangent is **`x - 2*sqrt 2*y + 6 = 0`

f(x) = y = sqrt(x+4)

Thus, at x = -2, f(-2) = y = sqrt(2)

Now, dy/dx = f'(x) = 1/{2*sqrt(x+4)}

f'(-2) = slope of the tangent at x = -2 = 1/{2*sqrt(2)}

Now, equation of the tangent at point [-2,1/{2*sqrt(x+4)}]

y - sqrt(2) = [1/{2*sqrt(2)}]*{x - (-2)}

or, y - sqrt(2) = [1/{2*sqrt(2)}]*{x + 2)}

or, 2y*sqrt(2) - 2*2 = x+2

or, 2y*sqrt(2) = x + 6