Verified answer

The derivative

y' = 15x^2 - 6

gives the slope of the tangent to the curve at any given point. At (-1,1),

y' = 15 (-1)^2 - 6 = 15-6 = 9

Having a point and a slope, we can write a point-slope equation in the form

y - b = m (x - a)

where m=9 is the slope and (a,b) = (-1,1) is the point:

y - 1 = 9 [x - (-1)]

y = 9x + 10 [slope-intercept form]

-9x + y = 10 [standard form]

First locate the spinoff of y i.e. y' = 15x^2 - 4 The slope of the tangent line on the factor (-a million.-) could be gained via calculating y'(-a million) i,e, y'(-a million) = 15(-a million)^2 - 4 = 15 - 4 = 11 subsequently the slope is 11 on the factor (-a million,-a million). utilising the factor-slope kind of the equation of a quickly line we've y - (-a million) = 11(x-(-a million)) -->............y = 11x + 11 - a million -->............y = 11x + 10