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

Saturday, January 24, 2015

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.

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