Equations of all the tangents..

[Calc12]

Find the equations of all the tangents to the graph of f(x) = x^2 - 4x + 25 that passes through the origin.


I've done this so far:

f(x) = x^2 - 4x + 25
f'(x) = 2x - 4

f'(0) = 2(0) - 4
= -4

f(x) = x^2 - 4x + 25
f(0) = 0^2 - 4(0) + 25
= 25

(0,25)

y=mx+b
25 =-4(0_ + b
b= 25

y= -4x + 25

Am I right? But how do you find ALL equations?

Thanks so much

 

[galactus]

The problem says the line(s) pass through the origin. Thus, the y intercept is 0.

So, the equations will have the form y=ax

Using \(\displaystyle y-y_{1}=m(x-x_{1})\), we can enter in the given info and solve for x. x is the x coordinate of the point(s) of tangency.

\(\displaystyle f(x)=x^{2}-4x+25\)

\(\displaystyle m=f'(x)=2x-4\)

Th line(s) pass through (0,0)

Enter them into the point slope form above:

\(\displaystyle \underbrace{x^{2}-4x+25}_{\text{y}}-0=\underbrace{(2x-4)}_{\text{m}}(x-0)\)

Solve for x. Then, sub this value back into f(x) to find the corresponding y coordinate.

These (x,y) will be the points of tangency. Use them and (0,0) to find the line equations.

 

[JeffM]

Let's start at the beginning. Did you sketch a graph?

If you do, you will SEE how many solutions there are. I know a geometric visualization is not a proof or a solution, but it is frequently very helpful in letting you know in what general direction you should be going when confronted with a new kind of problem.

So what does your vision say about the number of solutions?

There are an infinite number of tangents, and an infinite number of them do not run through the origin.

Can the relevant tangents also be described by equations other than the derivative of f(x)?

If so, let's call one of the alternative equations g(x). Let's say g(x) and f(x) are tangent to each other at x = t. Does that give you a hint?

 

[Calc12]

k so I got x = +/- 5

Now i sub +/- 5 into f(x)..

I get (5,30) and (-5,70)

Now i use y=mx + b?

& I get... the equations of the tangents are: y= 5+30x and y=-5+70x?

 

[JeffM]

Find the equations of all the tangents to the graph of f(x) = x^2 - 4x + 25 that passes through the origin.

k so I got x = +/- 5

Now i sub +/- 5 into f(x)..

I get (5,30) and (-5,70)

Now i use y=mx + b?

& I get... the equations of the tangents are: y= 5+30x and y=-5+70x?

You do?

At x = 0, y = 5 + 30x = 5. And at x = 0, y = - 5 + 70x = - 5. In my dotage as I am, those do not look like the origin to me.

I promised galactus that I would no longer respond to calculus problems because my mind is too old and funny. But this is no longer a calculus problem.
g(5) = f(5); g`(5) = f`(5), g(0) = 0 = h(0), h(- 5) = f(- 5), and h`(5) = f`(- 5), where g(x) and h(x) are the equations of the two required tangent lines. It helps if you set up separate functions for the two tangent lines, WHICH ARE LINEAR, and realize f(0) is irrelevant.

 

[Calc12]

JeffM,

when I graph y= 5+30x and y=-5+70x. they actually don't appear as tangents to the graph.

But If I sub +/- 5 into f'(x), i get y=6x and y=14x - they are tangents to the graph.

Guess that was the mistake I made?

 

[JeffM]

JeffM,

when I graph y= 5+30x and y=-5+70x. they actually don't appear as tangents to the graph.

But If I sub +/- 5 into f'(x), i get y=6x and y=14x - they are tangents to the graph.

Guess that was the mistake I made?

You have to understand that I am no mathmatician: I was a history major back in the day.

Where I think you went wrong was NOT to recognize that you were being asked to determine a different function of x from f(x). By keeping the f(x) notation, you missed the essential clue. Assuming there was at least one tangent that satisfied the reguirements and that tangent was the linear equation g(x), there was some t such that g(t) = f(t), g(0) = 0, and g`(t) = f`(t). You now solve for t. It turns out to be a quadratic, so it gives you 2 values, which would be consistent with your sketch. Galactus gave you another clue when he pointed out that a linear equation running through the origin has no constant term. But because he did not distinguish as clearly as he might have between the function whose tangent was being taken and the linear function running through the origin that was coincident with the tangent of the function at one unknown point, I suspect you missed his clue.

This is probably incoherent. What is tricky about this problem is that the derivative gives you a general function for the tangent at all points, but this question required you to find the point where the tangent met a specific constraint. For clarity's sake, the constraint needs to be treated as a separate function. I am sure galactus will (gently I pray) correct me if I am wrong.

Hope this helps.

 

[galactus]

k so I got x = +/- 5

Now i sub +/- 5 into f(x)..

I get (5,30) and (-5,70)

Now i use y=mx + b?

& I get... the equations of the tangents are: y= 5+30x and y=-5+70x?

You are not heeding what I mentioned in my first post. See your equation y=30x+5?. This says you have a slope of 30 and it crosses the y axis at y=5.

the problem says the line must cross at y=0. As I said before, there will be no b in your y=mx+b because it is 0...the origin.

You correctly found the points of tangency. (-5,70) and (5,30)

Now, you have (-5,70) and (0,0). Use these two points to find the line equation. You also have (5,30) and (0,0). Use these two points to find the line equation.

The line equations will have the form y=mx. No b because it is 0.

For instance, \(\displaystyle m=\frac{70-0}{-5-0}=-14\)

Your line equation is \(\displaystyle y=-14x\).

Now, find the other and you're done. Look at the graph. That helps determine what you're aiming for.

 

[soroban]

Hello, Calc12!

Let me give it a try . . .

\(\displaystyle \text{Find the equations of all the tangents to the graph of }f(x) \:=\: x^2 - 4x + 25\text{ that passes through the origin.}\)

\(\displaystyle \text{The tangent contains the Origin; its equation is: }\:y \,=\,mx\)


\(\displaystyle \text{It intersects the parabola at ONE point.}\)

. . \(\displaystyle x^2 - 4x + 25 \:=\:mx \quad\Rightarrow\quad x^2 - (m+4)x + 25 \:=\:0\)


\(\displaystyle \text{If the quadratic equation has one root, its discriminant is zero.}\)

. . \(\displaystyle (m+4)^2 - 4(25) \:=\:0 \quad\Rightarrow\quad m^2 + 8m - 84 \:=\:0\)

. . \(\displaystyle (m-6)(m+14) \:=\:0 \quad\Rightarrow\quad m \:=\:-14,\:6\)


\(\displaystyle \text{Therefore, the tangents are: }\:\begin{Bmatrix}y &=& 6x \\ y &=& \text{-}14x \end{Bmatrix}\)


\(\displaystyle \text{FYI: the points of tengency are: }\5,\,30),\;(\text{-}5,\,70)\)

 

[galactus]

Very nice, Soroban. I like that method better than mine.