find an equation in STANDARD FORM of the tangent line to the function at the given point

Question

1)    y=1-x-x^2-x^3-x^4 at x=-2

 

 

2)    y=2x(3x+2)^2 at x=-1

 

 

3)   y=(3x^4-6x^2+10)/ (12-2x^2-3x^3) at x=-1

 

This is what i got but i wasn't positive: 

 

1)  -23x+y=37

 

2) -14x+y=2

 

3) -35x+169y=126

 

Answer 1

1) y=1-x-x^2-x^3-x^4 at x=-2

 

y(-2)=1+2-4+8-16=-9

 

take the derivative of y with respect to x 

 

y'=-1-2x-3x^2-4x^3

 

y'(-2)=-1+4-12+32=23

 

The line tangent to function y at x=-2 contains the point (-2,-9) and has a slope of 23

 

23=(y+9)/(x+2)

 

you will need to rewrite this in the the standard

 

y=mx+b

 

we already know m is 23

Answer 2

1).  y = 1 - x - x² - x³ - x⁴

 

       When x = -2, y = 1 + 2 - 4 + 8 - 16 = -9.  So, (-2, -9) is a point          on the line.

 

      y' = -1 - 2x - 3x² - 4x³

 

      when x = -2, y' = -1 + 4 - 12 + 32 = 23 = slope of tangent line

 

      equation of tangent line:  y - (-9) = 23[x - (-2)]           

                                                   y + 9 = 23x + 46

                                                  23x - y = -37

 

2).  y = 2x(3x + 2)²

 

      When x = -1, y = -2(-1)² = -2.  So, (-1, -2) is on the line.  

 

      y' = 2(3x + 2)² + 2x(2)(3x + 2)(3)

 

      When x = -1, y' = 2(-1)² + -2(2)(-1)(3) = 2 + 12 = 14

 

      Equation of tangent line:  y - (-2) = 14[x - (-1)]

                                                   y + 2 = 14x + 14

                                                  14x - y = -12

 

3).  y = (3x⁴ - 6x² + 10)/(12 - 2x² - 3x³)

 

       When x = -1, y = 7/13.  So, (-1, 7/13) is on the line.

 

       y' = (12 - 2x² - 3x³)(12x³ - 12x) - (3x⁴ - 6x² + 10)(-4x - 9x²)

                                               (12 - 2x² - 3x³)²

 

      When x = -1, y' = (13)(0) - (7)(-5)    = 35 

                                               169                      169

 

      Equation of tangent line:  y - 7/13 = (35/169)[x - (-1)]

 

      Multiply by 169:  169y - 91 = 35(x + 1)

                                    169y - 91 = 35x + 35

                                    35x - 169y = -126

 

My answers to 1) and 3) are equivalent to yours.  

 

 

 

 

                 

                                                  

Answer 3

I will go over the first problem with you because the rest are similar.

 

First, evaluate y at   x = -2.

 

y = 1 - x - x² - x³ - x⁴

y = 1 - (-2) - (-2)² - (-2)³ - (-2)⁴

y = 1 + 2 - 4 + 8 - 16

y = -9

 

Your coordinate will be (-2, -9).  We will use this coordinate later on when finding the equation of the tangent line.

 

 

Second, we will take the derivative of y to find the slope of the tangent line.

 

y' = -1 - 2x - 3x² - 4x³

 

Evaluate y' at   x = -2.

 

y' = -1 - 2(-2) - 3(-2)² - 4(-2)³

y' = -1 + 4  - 12  +  32

y' = 23

 

 

Next we will find the equation of the tangent line using the point-slope form.

 

y - y₁ = m(x - x₁)      where m is slope, and (x₁ , y₁) are the coordinates points from step 1.

 

y - (-9) = 23(x - (-2))

y + 9 = 23(x + 2)

y = 23(x + 2) - 9

y = 23x + 46 - 9

y = 23x + 37

 

The equation of the tangent line is

 

y = 23x + 37

 

In standard form

 

-23x + y = 37