How do I find the tangent line of (x^4+3)^1/2 at (-1,2)

I'm having trouble with this question.

Steps would be awesome, and thanks in advance!

y = √(x^4+3)

y' = 4x^3/2√(x^4+3)
y' = 2x^3/√(x^4+3)

y'(-1) = -2/√4
y'(-1) = -1

So the slope is -1

y - 2 = -(x + 1)

First, confirm that the point is on the curve. If it is not, then you can stop there.
LS = ([-1]^4+3)^(1/2) = (1+3)^(1/2) = 4^(1/2) = +-2
RS = 2
So the point is on the curve. Keep going.
Determine the derivative of the given curve.
(x^4+3)^(1/2)
We need the power rule and the chain rule.
Chain rule: Take the derivative of the outside times the derivative of the inside. Keep going until you run out of insides.
Power rule: d ax^n / dx = anx^(n-1)
d (x^4+3)^(1/2) / dx
= (1/2)(x^4+3)^(-1/2)(4x^3)
= 4x^3 / 2(x^4+3)^(1/2)
At the point of tangency, x = -1
m = 4(-1)^3 / 2([-1]^4+3)^(1/2)
m = -4 / 2(4)^2
m = -4 / 2(16)
m = -1 / 8
So now, we can use point-slope form to give the equation.
y - y1 = m(x-x1), where (x1, y1) is a point on the line and m is the slope.
y - 2 = (-1/8)(x- -1)
y - 2 = (-1/8)(x+1)
Solve for y and simplify to express in slope-intercept form.