Slope of a curve question

Question

What are two examples of a function in which the line y=2x+1 is a tangent at some point on the curve.

Answer 1

Hi Anne,

The equation "y = 2x + 1" gives you a bit of information about the function you're looking for. You know right away that the slope of this tangent line is 2, so you need to find a function whose derivative is 2 at a certain point. You also know that the point has to follow the formula (x, 2x + 1).

You also know what the y-coordinate is at any value of x. Pick a random x value to evaluate, like x = 1.

y(1) = 2(1) + 1 = 3

So you need a function that has coordinate (1,3) and has a derivative of 2 at x = 1.

Let's look at

y = x^2

The derivative is

y' = 2x

and y'(1) = 2(1) = 2

So we've got the derivative part settled. Now we need to check if y(1) = 3

y(1) = (1)^2 = 1

Not quite. We're off by 2.

If we were to update

y = x^2

to

y = x^2 + 2

we'd be in good shape. The derivative would still be

y' = 2x,

and y(1) would actually be 3. Therefore:

y = x^2 + 2

does indeed have a tangent line of

y = 2x + 1.

Your question asked for two examples. I don't know if your teacher would appreciate or roll their eyes at the answer, but technically the function

y = 2x + 1

has a tangent line of

y = 2x + 1

at x = 1.

If you're worried you won't be able to get away with an example like that, you can use the same method above to find another equation that meets the criteria of

f'(x) = 2

and

(x, 2x+1)

Instead of using x = 1, let's try instead to use x = 2.

Since the tangent line is

y = 2x + 1

We need our function to have the point:

y(2) = 2(2) + 1 = 5

P(2,5)

For example,

y = 1/16*x⁴+4

y(2) = 1/16*(2⁴) + 4 = 5

P(2,5)

We also need the derivative to be 2 at x = 2

y'(x) = 1/4*x^3

y'(2) = 1/4*(2^3) = 1/4*8 = 2

It all checks out. The function

y = 1/16*x^4+4

has a tangent line of

y = 2x + 1

at x = 2.

Hope this helps!

Answer 2

Hi Anne B.,

A fine question, compliments to your instructor.

So what do you know already about integrals and differentials? I'm guessing, you know about them, BUT you aren't quite ready to apply that knowledge.

There are an infinite number of functions which are tangent to that line. So let's limit out search to just parabolas, which are the simplest type -- they curve away from a straight line tangent. And think: for the general parabola form y = ax^2 + b x + c , that parabola may be either concave upwards (a >0) and tangency therefore on to the right of the vertex, or concave downwards (a < 0), and tangency therefore to the left of the vertex. That's just a reminder, to consider global possibilities later!

So, that parabola has a slope function of y ' = 2a + b , which must match 2 (the slope of y = 2x+1 tangent line!). So that gives you b = 2(1-a) as one working equation.

The second working equation, you get from the actual point match : y = ax^2 + bx + c = 2x + 1

So when you substitute the first working equation into the second, you get:

ax^2 + 2(1-a)x + c = 2x + 1 .

You should immediately say, wait, I can't solve that uniquely for the roots of x, because I have two unknowns (a and c )!?! And so you do; if you pick specific values for a and c, though, you can solve through the quadratic for real roots for x, if you want, to get the x-axis intercepts of the parabola (to visualize it). Then solve the first working equation for b, and write the parabola equation.

Note that not all sets of a, b, and c will afford viable parabolae! The two x-axis intercepts of the parabola must be on the same side (either right, or left) of the intercept of the tangent line, I think you can geometrically see why?

-- Cheers, --Mr. d.