find the equation of tangent line to the curve y=3x²-4

Question

that is parallel to the line 3x+y=4.

Answer 1

The curve is y = 3x ^2 - 4, and , the tangent line is parallel to 3x + y = 4.

So parallel lines have same slopes.

Write the equation 3x + y = 4 in slope intercept form y = mx + c, where m is slope and c is y - intercept.

y = - 3x + 4.

Compare the above equation with y = mx + c.

Slope  is - 3.

Therefore slope of the tangent line is also - 3.

The implicit(or total) derivation of y = 3x ^2 - 4 is,

y ' = 6x.

Here y ' is slope of the tangent line.

So, 6x = - 3

⇒ x = - 3 / 6 = - 1 /2.

Using the x - value find the corresponding y - value with the curve.

y = 3(- 1 / 2)^2 - 4

= ( 3 / 4 ) - 4

= ( 3 - 16 ) / 4

⇒ y =  - 13 / 4.

Therefore the point is (- 1/2, - 13/4).

To find the tangent line equation, substitute the values of m = - 3 and (x, y ) = (- 1/2, - 13/4)  in the slope intercept form of an equation.

y = mx + c.

- 13 / 4 = (- 3)( - 1 / 2) + c.

c = (- 13 / 4) - 3 / 2

c = ( - 13 - 6) / 4

c = - 19 / 4.

Substitute m = - 3 and c in y = mx + c.

y = ( - 3 )x - ( 19 / 4 ).

The tangent line equation is y = - 3x - (19/4).