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

Answer 1

The tangent to a curve is a straight line that touches the curve at a certain point

and has exactly the same slope as the curve at that point.

The derivative of a function is the slope of the function at a certain point,

and so the tangent line to the curve.

The curve y  = 3x ² - 4

dy /dx  = 6x

The line 3x + y  = 4

y  = -3x  + 4

m  = -3

We know that parallel lines have equal slope to each other.

Tangent line slope is also -3.

m  = dy /dx

6x = -3

x = -1/2

Now find the values for y  at x = -1/2

y  = 3x ² - 4

y  = 3(-1/2) ² - 4

y  = (3/4) - 4

y  = (3 - 16)/4

y  = -13/4

Now we just need the line with slope of - 3 and that passes through the point (-1/2 , -13/4)

Line equation in point slope form y  - y 1 = m (x  - x1)

y - (-13/4) = -3 (x - (-1/2))

y  + 13/4 = -3 (x + 1/2)

y  + 13/4 = -3x  - 3/2

y  = - 3x  - 3/2 - 13/4

y  = - 3x - (6 + 13)/4

y  = - 3x  - 19/4

Equation of tangengent line is y  = - 3x  - 4.75.