How do you find the tangent line to the curve y=x^3-9xy=x3−9x at the point where x=1x=1?

2 Answers
Nam D.
Apr 21, 2018

y=-6x-2y=−6x−2

Explanation:

Given: y=x^3-9xy=x3−9x.

Let f(x)=y=x^3-9xf(x)=y=x3−9x. At x=1x=1, f(x)=1^3-9*1=1-9=-8f(x)=13−9⋅1=1−9=−8.

So, the point we are targeting is (1,-8)(1,−8).

To find the slope of the tangent line there, we must differentiate f(x)f(x) and then plug in x=1x=1 there.

:.f'(x)=3x^2-9

At x=1,

f'(1)=3*1^2-9

=3-9

=-6

So, the slope of the tangent line is -6.

Now, we use the point-slope formula to compute the equation, that is,

y-y_0=m(x-x_0)

  • (x_0,y_0) are the original coordinates

Therefore, we get,

y-(-8)=-6(x-1)

y+8=-6(x-1)

y+8=-6x+6

y=-6x+6-8

=-6x-2

A graph shows it:
desmos.comdesmos.com

Jim G.
Apr 21, 2018

y=-6x-2

Explanation:

•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = 1"

rArrdy/dx=3x^2-9

rArrdy/dx(x=1)=3-9=-6

rArry(x=1)=1-9=-8rArr(1,-8)

"using "m=-6" and "(x_1,y_1)=(1,-8)

y+8=-6(x-1)

rArry=-6x-2larrcolor(red)"equation of tangent"

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