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

Find an equation of the tangent line to the given curve at the specified point.

y= (x^2-1)/(x^2+x+1)   point=(1,0) Given the curve:

y=\frac{x^2-1}{x^2+x+1}

We have to find an equation of the tangent line to the given curve at the specified point (1,0).

Let us find the slope of the tangent by taking the first derivative of the curve.

y'=\frac{(x^2+x+1)\frac{d}{dx}(x^2-1)-(x^2-1)\frac{d}{dx}(x^2+x+1)}{(x^2+x+1)^2}

=\frac{2x(x^2+x+1)-(x^2-1)(2x+1)}{(x^2+x+1)^2}

=\frac{2x^3+2x^2+2x-2x^3-x^2+2x+1}{(x^2+x+1)^2}

=\frac{x^2+4x+1}{(x^2+x+1)^2}

Now the slope of...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Given the curve:

y=\frac{x^2-1}{x^2+x+1}

We have to find an equation of the tangent line to the given curve at the specified point (1,0).

Let us find the slope of the tangent by taking the first derivative of the curve.

y'=\frac{(x^2+x+1)\frac{d}{dx}(x^2-1)-(x^2-1)\frac{d}{dx}(x^2+x+1)}{(x^2+x+1)^2}

=\frac{2x(x^2+x+1)-(x^2-1)(2x+1)}{(x^2+x+1)^2}

=\frac{2x^3+2x^2+2x-2x^3-x^2+2x+1}{(x^2+x+1)^2}

=\frac{x^2+4x+1}{(x^2+x+1)^2}

Now the slope of the tangent at point (1,0) is:

y'=m=\frac{2}{3}

The equation of the tangent line can now be written as,

y-0=m(x-1)

y-0=\frac{2}{3}(x-1)

y=\frac{2}{3}x-\frac{2}{3}

Approved by eNotes Editorial Team The general equation of a line passing through the point (a, b) and with slope m is (y - b)/(x - a) = m.

We need to determine the equation of a line that passes through the point=(1,0). The value of (a, b) here is (1,0).

Given a curve y = f(x), the slope of a tangent at the point (p, q) is f'(p).

For the curve y= (x^2-1)/(x^2+x+1)

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

= ((x^2 - 1)'(x^2+x+1) - (x^2-1)*(x^2+x+1)')/(x^2+x+1)^2

= ((2*x)'(x^2+x+1) - (x^2-1)*(2*x + 1))/(x^2+x+1)^2

= ((2*x)'(x^2+x+1) - (x^2-1)*(2*x + 1))/(x^2+x+1)^2

At the point (1,0) the value of y'=((2*1)'(1^2+1+1) - (1^2-1)*(2*1 + 1))/(1^2+1+1)^2

= (2*3 - 0)/3^2

= 2/3

Now the equation of a line at a tangent to (1,0) the curve y= (x^2-1)/(x^2+x+1)   and passing through (1,0) is

(y - 0)/(x -1 ) = 2/3

3y = 2x - 2

2x - 3y - 2 = 0

The equation of the required line is 2x - 3y - 2 = 0

In the graph below, the given slope is in black and the tangent is in red.

Approved by eNotes Editorial Team The slope of the line tangent to the curve at a given point is the value of the derivative of the function describing the curve at this point.

The derivative of y(x) = (x^2 - 1)/(x^2 + x + 1) can be found using the quotient rule:

(f/g)' = (f'g - fg')/g^2

Here, f = x^2 - 1 and g = x^2 + x + 1

f' = 2x  and g' = 2x + 1

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

Simplify by multiplying out the parenthesis:

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

y'(x) = (x^2+4x+1)/(x^2 + x + 1)^2

At the point (1, 0), the slope of the tangent line will then be

m = y'(1) = (1^2 + 4*1 + 1)/(1^2 + 1 + 1)^2= 6/9 = 2/3

The equation of the line passing through (1,0) in point-slope form will be

y - 0 = 2/3(x - 1)

In the slope-intercept form, the equation of the tangent line to the given curve at the point (1, 0) is

y = 2/3x - 2/3

Approved by eNotes Editorial Team 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

Approved by eNotes Editorial Team