Not-so-easy tangent line

Question

Find the equation of the line that is tangent to y = x^2 - 5x + 6 and passes through the point (0, -3).

Answer 1

We know that some point on the curve, y = x^2 - 5x + 6 -- let’s  say (x₀, y₀) -- has a tangent line that passes through(0, -3). Note that (0,-3) is not on the curve, since it does not satisfy the equation of the curve.

We do have two ways in which we can describe the slope of this tangent line:

1)     The slope of the line contains two points: (0, -3) and (x₀, y₀). So the slope of this line can be written as

m = y₀ – (-3)  = y₀ + (3)  

         x₀ -  0              x₀

2)     The slope of the line at any point on the curve can also be represented as the value of the derivative at that point:

f(x) = x^2 - 5x + 6 

f’(x) = 2x - 5

f’(x₀) = 2x₀ - 5

Therefore, we can set the two representations equal to each other

 y₀ + (3)   = 2x₀ - 5

       x₀

Now express y₀ in terms of x₀, and solve the resulting  equation

x₀^2 - 5x₀ + 6  + 3  = 2x₀^2 - 5x₀

                             9  =  x₀^2

                              so           x₀ = +- 3

Substituting x₀ = 3, into the derivative  yields m = f’(3) = 1 so    y = x - 3 is one solution

Substituting x₀ =  -3 yields m = f’( -3) = -11, so  y = -11x - 3 is the other solution