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

The warning in the problem is the whole key, so it is worth starting there. Normally you find a tangent line by putting the given x into the derivative, but that only works when the point is actually on the curve. Here P(1, 0) is not on it, since f(1) = 2 and not 0. So f'(1) = 2 is not the slope of anything you are looking for, and that is the first thing most people write down.

What makes this work is treating the point of tangency as unknown. Call it (a, a² + 1), since every point on the curve has that form.

The slope of the tangent there can now be written two different ways, and setting them equal is what pins down a. From the derivative, f'(x) = 2x, so at x = a:

slope = 2a

And from the two points the line runs through, (a, a² + 1) and (1, 0), the slope is just rise over run:

slope = (a² + 1 − 0) / (a − 1)

Setting those equal and clearing the fraction:

2a(a − 1) = a² + 1

2a² − 2a = a² + 1

a² − 2a − 1 = 0

That one does not factor, so complete the square:

(a − 1)² = 2

a = 1 ± √2

Two values of a, so there are two tangent lines, which should feel right. From a point sitting below a parabola you can always draw exactly two tangents to it.

Now finish the job, because the question asks for the equations of the lines and not just where they touch. Each slope is 2a, so the slopes are 2 + 2√2 and 2 − 2√2, and both lines pass through (1, 0), which makes point slope form the easy way to write them:

y = (2 + 2√2)(x − 1)

y = (2 − 2√2)(x − 1)

In decimals those slopes are about 4.83 and −0.83, which matches the picture: one tangent climbs steeply up the right side of the parabola, and the other tips just slightly downward to the left.

I checked both of these by subtracting each line from x² + 1 and confirming the quadratic left over has a discriminant of exactly 0. That is what tangency means, the line meeting the curve at one point instead of cutting through it at two, so it is a good way to be sure you have not just found a secant line.

So, our final answer is y = (2 + 2√2)(x − 1) and y = (2 − 2√2)(x − 1).

Answer 3

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)