Find the equation for the tangent line to the function f(x) = 4x3+3x2+5x+6 when x = 2. put your answer in slope intercept (y= mx+ b.format, and use it to identify the y-intercept for this line. the y-intercept for the tangent line to the function f(x) = 4x3+3x2+5x+6 when x = 2 is

The derivative of the function at x=2 is

... f'(x) = 12x² +6x +5

... f'(2) = (12·2 +6)·2 +5 = 65

The value of the function at x=2 is

... f(2) = ((4·2 +3)·2 +5)·2 +6 = 60

Then the point-slope form of the line can be written as

... y = 65(x -2) +60

... y =  65x -70 . . . . . . eliminate parentheses, collect terms

The y-intercept of the tangent line at x=2 is -70.

x=-36

Step-by-step explanation:

Add the 7/2x +1/2x to be 8/2x which means that it'll be 4x=18=9/2

then subtract 9/2x on both sides so itll be

-9/2x       -9/2x

  because if you do 8/2-9/2 it's -1/2

then multiply the reciprocal to both sides so

which then cancels out -1/2 and makes it

x=-36