Question

How do you find the equations of the tangent to the curve `y = (1 - x)(1 + x^2)-1` that pass through the point (1, 2)?

Answer 1

Find the equation for the line tangent to the curve at `x=a`. Then substitute `x=1` and `y=2` to find `a`. The rewrite the tangent line using that value of `a`. I got `y = -7.95x+9.95` (approx).
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Explanation:

Simplify the function `y = f(x) = -x^3+x^2-x`

We shall find the equation of the line tangent to the curve at `(a,f(a))`.

`f(a) = -a^3+a^2-a`

`f'(x) = -3x^2+2x-1`, so the slope of the tangent line is

`m = f'(a) = -3a^2+2a-1`

The tangent line has equation:

`y-(-a^3+a^2-a) = (-3a^2+2a-1)(x-a)`

The slope-intercept form is

`y = (-3a^2+2a-1)x+(2a^3-a^2)`

we shall now find `a` to make the tangent line contain the point `(1,2)`

Solve for `a`

`2 = (-3a^2+2a-1)(1)+(2a^3-a^2)`

`2a^3-4a^2+2a-3 = 0`

Use whatever tools you have to solve this cubic. I got

`a ~~ 1.89`

So the equation of the tangent line is (approximately)

`y = -7.95x+9.95`