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

Answer 1

#y=-6x-2#

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:

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

#y=-6x-2#

#•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"#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To find the tangent line to the curve y=x^3-9x at the point where x=1, we need to find the slope of the curve at that point.

First, we find the derivative of the function y=x^3-9x with respect to x. The derivative of x^3 is 3x^2, and the derivative of -9x is -9. Therefore, the derivative of the function is dy/dx = 3x^2 - 9.

Next, we substitute x=1 into the derivative to find the slope at that point. dy/dx = 3(1)^2 - 9 = -6.

The slope of the tangent line is -6.

To find the equation of the tangent line, we use the point-slope form of a line. The equation is y - y1 = m(x - x1), where (x1, y1) is the point on the curve where x=1, and m is the slope.

Substituting x=1 and y=1^3-9(1) into the equation, we have y - (-8) = -6(x - 1).

Simplifying, we get y + 8 = -6x + 6.

Rearranging the equation, we have y = -6x - 2.

Therefore, the equation of the tangent line to the curve y=x^3-9x at the point where x=1 is y = -6x - 2.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7