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

2 Answers
Apr 21, 2018

Answer:

#y=-6x-2#

Explanation:

Given: #y=x^3-9x#.

Let #f(x)=y=x^3-9x#. At #x=1#, #f(x)=1^3-9*1=1-9=-8#.

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

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

Apr 21, 2018

Answer:

#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"#