find the equation of tangent line to the curve y=3x²-4 that is parallel to the line 3x+y=4.

Question

i dont know the answer

Answer 1

dy/dx=6x the tangent, gradient or slope at point (x,3x^2-4). The slope of the line 3x+y=4 or y=-3x+4 is -3. So we need the tangent to be -3 in order to be parallel (same slope). Therefore, 6x=-3 and x=-1/2, making y=3/4-4=-13/4. The point of the tangent is (-1/2,-13/4) and the equation of the tangent must pass through this point. y=mx+c is the general equation of a line with gradient m, so m=-3. -13/4=-3*(-1/2)+c; -13/4=3/2+c; c=-13/4-3/2=-19/4 and the equation of the tangent is y=-3x-19/4, or 4y+12x+19=0.