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

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### 3 Answers

We have to find the tangent to the curve y = x^2 + 8x + 12, at the point where x= -4.

Now the slope of the tangent to a curve is given by the derivative of the equation representing the curve. In the case of y = x^2 + 8x + 12, the derivative is

y' = 2x + 8

At the point where x = -4,

y' = 2*(-4) + 8 = -8 + 8 = 0.

So the slope of the tangent is 0.

Now substitute x = -4 in the equation y = x^2 + 8x + 12

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

=> y = 16 - 32 + 12

=> y = -4

**Therefore the equation of the tangent to the curve at the point where x = -4 is y = -4.**

The equation of tangent at (x1,y1)for a curve y = f(x) is given by ;

y - y1 = f'(x1) (x-x1).

The given function is f(x) = x^2+8x+13.

At x1= -4 , y1 = (-4)^2+8*(-4)+12 = 16 -32+12 = -4.

f(x) = (x^2+8x+12) = 2x+8.

f'(x1) = f(-4) = 2(-4) +8 = 0.

Therefore the equation of the tangent.

Therefore the tanget at (-4, -4) is given by:

y - 4 = 0(x--(4)).

Or y -4 = 0. Or y = 4 is the tangent at x= -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.**