Find An Equation Of The Tangent Line To The Given Curve At The Specified Point. Y = X2 − 1 X2 + X + 1 , (1, 0)

1 Answer

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

`y= (x^2-1)(x^2+x+1)`

First we will find the derivative of y.

==> We will use the quotient rule.

`==gt y' = ((x^2-1)'(x^2+x+1)- (x^2-1)(x^2+x+1)')/(x^2+x+1)^2 `

`==gt y' = (2x(x^2+x+1)- (2x+1)(x^2-1))/(x^2+x+1)^2 `

`==gt y'= (2x^3 +2x^2 + 2x - 2x^3 +2x +x^2 -1)/(x^2+x+1)^2 `

`==gt y' = (3x^2 + 4x -1)/(x^2+x+1)^2`

==> Now we subsitute with x = 1

`==gt y'(1)= (3+4-1)/(1+1+1)^2 = 6/9 = 2/3`

Then, the slope of the tangent line is m= `2/3`

Now we will find the equation of the tangent line where the slope is m= 2/3 and the point (1,0).

`==gt y-y1= m(x-x1) `

`==gt y= 2/3(x-1)`

`==gt y= 2/3 x - 2/3`