Finding the equations of two lines that are tangent to the curve and parallel to a line. I have no idea how to do this as i have been absent from class for medical reasons.
the curve $y=3+x^2$
and the line $3x-y=-6$
The answer is a possibility of the following:
a) $ y=3x, y=3x+4$
b) $ y=12x-15, y=12x+17$
c) $ y=-3x+9, y=-3x-16$
d) $y=12x-13, y=12+19$
e) $y=-12x-16, y=-12x+8$
f) $y=3x+5, y=3x+1$
 A: A line being tangent to a curve means that it "touches it" i.e. there is a shared point of the line and the curve, and at that point the gradients are the same. Two lines being parallel means that they have the same gradient.
What you need to do is find the general equation of any line parallel to the given one, then find the points of intersection of this general parallel line with the given quadratic curve. The lines that are tangent to the curve will satisfy the condition that, at the points where the lines intersect the curve, the gradient of the line will be the same as the gradient of the curve.
A: As said above, the slope of the required line is 3 but there will be only one line which is tangent to the curve.
The given curve is a parabola and there is only one tangent of a given slope touching it.
$y=3 +x^2$
$dy/dx = 2x$  
Since the slope of the line is 3
$dy/dx=2x=3$
Therefore, $x=1.5 $ and $y=3+(1.5)^2 =5.25$
Therefore the equation of given line will be $y=3x+0.75$
A: The equation of any line parallel to $3x-y=-6$ can be written as $3x-y=k$
Now, let this line touches the curve $y=3+x^2$ at $(a,b).$
So, $3a-b=k$ or $b=3a-k$  and $b=3+a^2$
Comparing the values of $b$ we get, $3a-k=3+a^2,$ or $a^2-3a+k+3=0$
Now, this is a quadratic equation in $a,$ for the tangency both the root should be same, i.e, discriminant$(-3)^2-4\cdot 1\cdot(k+3)$ must be $0\implies 9-4k-12=0,k=-\frac 3 4$ 
So, the equation of the tangent becomes  $3x-y=-\frac 3 4,$ or $12x-4y=-3$
