Calculus Help!!! I want to know how to work them out!

Q#1 The equation of the tangent to the curve y=x^2 - 4x at the point where the curve crosses the y-axis is:

(a) y=8x-4
(b) y=-4x
(c) y=-4
(d) y=4x
(e) y=4x-8

Q#2 Does the Mean Value Theorem apply to G(x) = |3x-4| on [1,2]

(a) yes it does
(b) no, G is not a function
(c) no, G is not continuous
(d) no, G isnot differentiable
(e) none of these

Q#3 The zeros of P(x)= (x^3 - x^2 - 4x +4) / (x-2) are:

(a) [1, 2, -2, 5/3]
(b) [1, 2, 3]
(c) [1, -2]
(d) [1]
(e) none of these

Q#4 If P(x) =
[x(x-2)] / (4 - x^2) , if x=2
k-1, if x=2
AND P(x) is continuous at x=2, then k=

(a) 1/3
(b) 3
(c) 1/2
(d) 2
(e) -1/2

1) First of all, when y = x^2 - 4x crosses the y-axis, we set x = 0:
==> The point is (0, 0).

Now, y' = 2x - 4
==> The slope of the tangent at x = 0 is y'(0) = -4.

So, the equation of the tangent line is y = -4x.
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2) Note that G has a corner point when 3x - 4 = 0 <==> x = 4/3, which is in [1, 2].
Hence, G is not differentiable on [1, 2] (and so MVT does not apply).
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3) Since x^3 - x^2 - 4x + 4
= x^2(x - 1) - 4(x - 1)
= (x^2 - 4)(x - 1)
= (x - 2)(x + 2)(x - 1),

P(x) = (x - 2)(x + 2)(x - 1)/(x - 2) = (x + 2)(x - 1) for x ≠ 2.
(Note that x = 2 is a removable discontinuity and not a zero.)

So, the zeros of P(x) are x = -2, 1.
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4) For continuity at x = 2, we need lim(x→2) f(x) = f(2).
==> lim(x→2) x(x - 2)/(4 - x^2) = k - 1
==> lim(x→2) x(x - 2)/[(2 + x)(2 - x)] = k - 1
==> lim(x→2) x(x - 2)/[(2 + x) * -(x - 2)] = k - 1
==> lim(x→2) -x/(2 + x) = k - 1
==> -1/2 = k - 1
==> k = 1/2.

I hope this helps!