draw the tangent line to the graph of y= x^3-12x at x=2?

Question

please show your solution. tnx a lot.

Answer 1

f'(x)=3x^2-12
3(x+2)(x-2)=0
slope is 0 at x=2
equation of the tangent is
y+16=0

Answer 2

First solve the equation for y at x=2.

y=(2)^3-12(2)=-16

You then have to find the derivative of the function, since the slope of the tangent line can be found by plugging x=2 into the derivative.

dy/dx=3x^2-12
dy/dx=3(2)^2-12=0

The tangent line has a slope of zero, so it is therefore a horizontal line.

Plug in the points (2,-16) into the point-slope formula along with the slope of the tangent line to be able to draw the tangent line.

y-(-16)=x(x-2)
y+16=0(x-2)
y+16=0
y=-16.....this is the equation of the tangent line at x=2.

Answer 3

the line x + 5 = 0 is a vertical line. so that you're looking for tangent lines perpendicular to this. they're going to be horizontal lines. Take the by-product of the function and set it equivalent to 0. The factors the position the by-product is 0 may have horizontal tangents. y = (x² - a million)² dy/dx = 2(x² - a million)(2x) = 4x(x² - a million) = 4x(x - a million)(x + a million) = 0 x = -a million, 0, a million fixing for y at each and every of those factors we've: (-a million,0), (0,a million), and (a million,0) So the tangents to the curve at those factors are: y = 0 and y = a million. note that the line y = 0 is tangent to 2 of the criteria.

Answer 4

First, find out what y equals by substituting x=2 into the original equation to get y= -16.
Second, find the derivative of the graph, so it is
y'= 3x^2-12=m(slope, because the derivative is the slope of the tangent line to a function).
Third, plug in x=2 into m to get m=0.
Plug the x, y and m values in y=mx+b, so you get
-16=0(2)+b=b. Therefore, the line is y=-16.

Answer 5

tangent lines have the same slope as the curve at the point if intersection
y=x^3-12x
y'=3x^2-12
y'(2)=3*2^2-12=0
so y=c
y=x^3-12x
y(2)=2^3-12x=8-12*2=-16
so
y=-16 is the answer.

Answer 6

when x = 2, y =-16
when x=0, y=0

gradient at x=2 is-8

thus, eqn of tangent at x=2,
y-(-16) = (-8)(x-2)

Answer 7

dy/dy = 3x^2-12
@x=2, f'(x) =0
f(x)=8-24=-16

y+16=0(x-2)
y = -16 for all x

Answer 8

are you even looking at your homework problems before you type them up here?

Answer 9

Ok, I drew it, what next?

Answer 10

already got enough answers