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

find lines tangent to y=x^2+1 that also go through (1,0)

it's very easy to make arithmetic and algebra mistakes in calculating the tangent lines

there are two of them, one upward sloping, one downward

y =-2x +4-2sqr2 and y =-2x+4+2sqr2

graph them to convince yourself it's accurate,

tangent lines will have slope = derivative of y = 2x

those lines go through (1,0) and a point on the curve (x, y) = (x, x^2+1)

it helps to draw at least a rough sketch of the upward opening parabola and plot the points

as there are mistakes below

the parabola has y intercept 1, and x intercepts 1 and -1, with vertex (0,1)

y = a(x-1)(x+1)

1 = a(-1)(1) = -a

a = -1 mistake somewhere as a negative "a" means a downward opening parabola

slope = 2x, = difference in y over difference in x = (x^2+1 - 0)/(x -1) = (x^2+1)/(x-1) = 2x

2x^2 -2x = x^2 +1

x^2 -1 -2x= 0

using quadratic formula

x = 1 + or - (1/2)sqr(4+4) = 1+/-sqr2 = about 2.414 or 0.414

y = x^2 +1 = (1+sqr2)^2 +1 = 1+2+2sqr2 +1 = 4+2sqr2

or y = x^2+1 = (1-sqr2)^2 +1 = 1 -2 -2sqr2 +1 = -2sqr2

line through (1,0) to (1+sqr2, 4+2sqr2) with slope = m = 2+2sqr2

two point formula

y -y1 = m(x-x1)

y -0 = (2+2sqr2)(x-1)

and a 2nd line with x=1-sqr2, y=-2sqr2

line from (1,0) to (1-sqr2, -2sqr2)

use two points formula again

Answer 2

Since we need tangent lines, we need the first derivative of x² + 1, which is 2x. This becomes the expression for the slope of any tangent line at point x. This gives us the line equation:

y - 0 = (2x)(x-1). (using point-slope form)

so y = 2x² - 2x

Now, to be a tangent line, it must also touch the graph of f(x), which means that there must be a point in the form (x, f(x)) or (x, x² + 1) that satisfies the equation. Subbing this in gives:

x² + 1 = 2x² - 2x

Solving the quadratic will give you the x-value(s) where this occurs. Sub in the x-values to your first derivative expression (2x) to get the slope(s) for the line(s) that are tangent and passes through (1, 0).

Answer 3

Point-slope form: y - y_p = m(x - x_p)

To get the slope of the tangent line, take the derivative of the function x² + 1 (which I will write as f'(x)). The equation for the tangent lines of x² + 1 going through the point (1, 0), using the formula for point-slope form of a line is

y - 0 = f'(x)*(x - 1)

y = f'(x)*(x - 1)

Since we know that the line must ALSO touch the function x² + 1, we can replace y in our tangent line equation with this function:

x² + 1 = f'(x)*(x - 1)

Solve for x. Your line equations should look like

y = mx - m

with m replaced with the derivatives at the x-values you solve for.