helpp!! :) Calculus

Question

Find the equations of all lines that are tangent to the graph of the function f(x) = x ^ 2 + 1 and pass through the point P(1,0). Be careful: this point is not on the graph of f.

Answer 1

Let P(x₁,x₁² + 1) be any point on the parabola. Since the derivative f'(x) = 2x, the slope at P is 2x₁.

For the tangent line(s) passing through P and (1,0) the slope can also be calculated as (x₁²+1 - 0) / ( x₁-1)

Setting the two expressions representing the slopes equal to each other gives:

2x₁(x₁-1)=x₁²+1

x₁² - 2x₁ - 1 = 0

By quadratic formula (or completing the square), x₁ can have values:

1+√2 OR 1-√2

The slopes of the lines at the points with those x-coordinates are respectively:

2(1+√2) and 2(1-√2)

Since the lines must pass through the point (1,0), the equations are:

y = 2(1+√2)(x-1)

and

y = 2(1-√2)(x-1)

desmos.com/calculator/ifbejpjbxx

Answer 2

let (a,a^2+1) be the tangency point(s),

then with y =x^2 +1, the slope = y' = 2x = 2a

= (y2-y1)/(x2-x1) = (a^2 +1 -0)/(a-1) = 2a

a^2 +1 = 2a^2 -2a

2a^2 -a^2 -2a -1 = 0

a^2 -2a =1

a^2 -2a +1 = 1+1

(a-1)^2 = 2

a-1 = +/-sqr2

a = 1 +/-sqr2 = x coordinate of the tangency points

a^2 +1 = (1+/-sqr2)^2 +1 = 4+/-2sqr2 = y coordinates of the tangency points

(1-sqr2, 4-2sqr2) and (1+sqr2, 4+2sqr2)

tangency lines are

one downward sloping y=(2-2sqr2)(x-1)

and

one upward sloping y = (2+2sqr2)(x-1)