matematikk.net

• En kurve
  y(x) = x2 + 1
  
  y = x2 + 1

• y'(x) = 2x

• y = mx + b
  5 = 4×2 + b
  b = -3

Here's a slightly more complicated variant on the same problem. Find the two lines that are tangent to

y = x2 - 2x + 1

and pass through the point, (5, 7). Observe that in this case, the point, (5, 7), does not lie on the curve. So you are finding lines that pass through a point outside the curve but at some other points, which are as yet unknown, they are tangent to the curve. The strategy is to identify those points of tangency. From them, it is easy to find the lines that solve the problem.

Step 1: Find the derivative of the curve. In this case we have

y' = 2x - 2

We do this because we know that at the point of tangency, the derivative of the curve must be equal to the slope of the tangent line. So we need to know this derivative.

Step 2: Write as much of the equation of the line as you can from what you know. The slope, m, of the line is still unknown. But you know what point it must pass through. In this case that is the point (5,7). So using the formula for the equation of a line passing through a given point with a given slope, you have:

y - 7 = m(x - 5)

Step 3: Use the derivative to write an equation for the slope. We don't know the slope of the line yet, but we know that if the point, (x,y), is the point of tangency, then the slope of the line will be equal to the derivative of the original curve at x. The derivative we already determined to be y' = 2x - 2. So we write the equation

m = y' = 2x - 2

Step 4: Substitute the slope expression into the equation for the line. You have m = 2x - 2 and y - 7 = m(x - 5). Put them both together by substituting for m in the second equation.

y - 7 = (2x - 2)(x - 5) = 2x2 - 12x + 10

y = 2x2 - 12x + 17

Step 5: Substitute the equation for the original curve back in. That is, you know that y = x2 - 2x + 1, where (x,y) is the point of tangency. Why? Because the point of tangency must lie on the curve. Substituting that expression for y into the above gives:

x2 - 2x + 1 = 2x2 - 12x + 17

Graph of solution

Step 6: Gather like terms and solve for x. Using simple algebra the above equation becomes the quadratic

x2 - 10x + 16 = 0

You can either use the quadratic formula, or you can factor this one in your head to get

(x - 2)(x - 8) = 0

So the x coordinate of the the point of tangency is either x = 2 or x = 8.

Step 7: Substitute back to get m. That is, we have already seen that m = 2x - 2 at any point, (x, y), of tangency. Since we now know what both the x's are for the points of tangency, we can put those values into the equation for m and get that either m = 2 or m = 14. From the equation we had in step 2,

y - 7 = 2(x - 5)

and

y - 7 = 14(x - 5)

When you multiply those out and put them into slope-intercept form, the equations of the two lines that are tangent to y = x2 - 2x + 1 and pass through the point, (5, 7), are

y = 2x - 3

and

y = 14x - 63

and you are done. The curve and the two lines are illustrated in the graph on the right here. Notice where the two lines intersect. Both points of tangency are visible on the graph, although one is very nearly at the top of the graph.