find the equation of the tangent line at x=c, y=(1/2)x^2-4, c=2?

Answer 1

y=(1/2)x^2-4
y'=x
at x=2,y=-2
equation
y+2=2(x-2)
y+2=2x-4
2x-y-6=0

Answer 2

I expect you know how to differentiate
(1/2)x^2-4

(Hope you get x as the answer!)

To find the gradient of the tangent, you have to substitute c for x in this derivative. So the gradient is 2.

Now you have to find the value of y when x = c, so substitute c (i.e. 2 in this case) for x in
(1/2)x^2-4

Now use the formula
y - y(1) = m*(x - x(1))

with m = 2 (we found),
x(1) = c (i.e. 2 in this case)
y(1) = the value of y we just found by substituting -- yes, I know I haven't worked it out, I'm sure you can do that yourself.

Answer 3

y = (1/2)x^2 - 4

You just find the derivative of the expression:

y' = (1/2)(2x) - 0

Now you can simplify the expression:

y' = x

Now, In order to find the equation of the tangent line, you must find any point that the curve goes through.