# Tangent.Find the equation of the tangent for y = x^2 + 8x + 12, where x = -4.

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You need to use the equation of tangent line to the graph, at a point `(x_0,y_0)` , such that:

`f(x) - f(x_0) = f'(x_0)(x - x_0)`

Since the problem provides `x_0 = 4` , you may find `f(x_0) ` substitutig 4 for `x_0` in equation of function, such that:

`f(4) = (-4)^2 + 8*4 + 12 => f(4) = 16 - 32 + 12 => f(-4) = -4`

You need to evaluate `f'(x_0)` , hence, you need to find derivative of the function at `x_0 = 4` , such that:

`f'(-4) = lim_(x -> -4) (f(x) - f(-4))/(x + 4)`

`f'(-4) = lim_(x -> -4) (x^2 + 8x + 16)/(x + 4) = 0/0`

Since the limit gives an indetermination `0/0` , you may use l'Hospital's theorem, such that:

`lim_(x -> -4) (x^2 + 8x + 16)/(x + 4)= lim_(x -> -4) ((x^2 + 8x + 16)')/((x + 4)') `

`lim_(x -> -4) (x^2 + 8x + 16)/(x + 4)= lim_(x -> -4) (2x + 8)/1`

`lim_(x -> -4) (2x + 8) = 2*(-4) + 8 = 0`

Hence, evaluating derivative of the function at `x = -4` yields` f'(4) = 0` .

Hence, you may write the equation of tangent line, such that:

`y - (-4) = 0*(x - 4) => y + 4 = 0 => y = -4`

**Hence, evaluating the tangent line to the given parabola, at `x = -4` , yields that it is the line `y = -4` .**

We'll have to find first the derivative of y = x^2 + 8x + 12, knowing that for any value of x, y' represents the slope of the tangent at that point.

We'll differentiate y:

y' = 2x + 8

For x = -4, we'll get:

y' = 2*(-4) + 8

y' = -8 + 8

y' = 0

If the slope is zero, the tangent to the curve is parallel to x axis.

We'll determine y for x = -4:

y = (-4)^2 + 8*(-4)+ 12

y = 16 - 32 + 12

y = -4

The equation of the tangent to the curve is y = -4.