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

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 2

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.