# Minimum veritcal distance between two graphs.

1. Sep 4, 2010

### hobomoe

1. The problem statement, all variables and given/known data
Calculate the minimum vertical distance between the cubic y=x^3-x^2-4x+4 and the parabola y=-2x^2+16x-30, when x is positive.

2. Relevant equations
y=x^3-x^2-4x+4
y=-2x^2+16x-30

3. The attempt at a solution
I have no idea at all how to work this out, so perhaps someone could walk me through it?

2. Sep 4, 2010

### hobomoe

Since the x intercepts of each graph closest to each other are 2 and 3, the line must cut through 2.5. The coordinates for graphs are now (2.5,y1) and (2.5,y2). Substituting 2.5 into the equations gives the the y coordinates as 3.375 and -2.5. No need to use 2.5 to calculate distance since the gradient doesn't exist, so 3.374+2.5=5.875.

Not sure if this is right, but if it is could someone tell me how to work it out in a more calculus kind of way since this doesn't seem at all like calculus to me.

3. Sep 4, 2010

### Mentallic

It's a nice attempt, but not exactly right. You can't make the assumption that since they're between 2 and 3 then it must be 2.5.
Take this graph, $$y=\frac{8x-16}{5-2x}$$. It's x-intercept is 2 but it has an asymptote at 2.5 so this is an extreme example of how it doesn't work.

If we have y=f(x) and y=g(x) then we let them equal each other so we have f(x)=g(x) then y=f(x)-g(x), we are finding x where the graphs of f(x) and g(x) are equal, or in other words, where f(x)-g(x)=0. But since in our case they don't cross, we are trying to find x where they are closest to each other. We are trying to find where y is a minimum.

4. Sep 4, 2010

### hobomoe

How do you find the x intercept?

5. Sep 4, 2010

### Mentallic

There is no x-intercept.
Again, in the function y=f(x)-g(x) where y is a minimum.

6. Sep 4, 2010

### hobomoe

Whats g(x)? Is it f(x) for the 2nd equation?

7. Sep 4, 2010

### Mentallic

Yep! Sorry I was assuming you knew about functions considering you're doing calculus already.

8. Sep 4, 2010

### hobomoe

I do, but we don't use g(x), only f(x). Could you show me the first steps of working to get this? I'm pretty confused still.

9. Sep 4, 2010

### Mentallic

Well, what's y=f(x)-g(x) ?

10. Sep 4, 2010

### hobomoe

x^3+x^2-20x+34?

11. Sep 4, 2010

### Mentallic

Right, so we have the graph y=x3+x2-20x+34. This graph tells us the distance between our function f(x) and g(x) since we are finding the difference between them. So at x=2, it will give us y=6 so that means the distance at x=2 between f(x) and g(x) is 6. How do you find where the minimum distance is?

12. Sep 4, 2010

### hobomoe

I have no idea :(

13. Sep 4, 2010

### Mentallic

Do you know calculus?

14. Sep 4, 2010

### hobomoe

Yea but I'm having a mental block at the moment.

15. Sep 4, 2010

### Mentallic

It's really simple! On the graph it tells us the distance between the two original curves f(x) and g(x). At x=2 we have y=6 which tells us the distance between the two curves is 6, at x=3 the distance between them is 10. Obviously the minimum distance between them is then the minimum point on the graph!
If I asked you to find the minimum of the parabola y=x2+2x using calculus, what would you do?

16. Sep 4, 2010

### hobomoe

EDIT: Never mind with that, bit late

17. Sep 4, 2010

### Mentallic

Don't think about finding what the actual distance is yet, we just want to find the x-value where the minimum occurs, which is where calculus comes into play. After we do this we'll substitute it back into equation to find the distance.

18. Sep 4, 2010

### hobomoe

Is the minimum the turning point, so I need to find when the gradient equals zero?

19. Sep 4, 2010

### Mentallic

You should be more confident in your abilities, rather:
"The minimum is the turning point, so I need to find when the gradient equals zero!"

:tongue:

20. Sep 4, 2010

### hobomoe

f'(x)=3x^2+2x-20
I can't seem to manage factorizing this now, although I may be going in the wrong direction.