# Homework Help: Intersection of two graphs

1. Nov 21, 2009

### TyErd

the graph of y=kx+3 intersects the graph of y=x^2 +8x at two distinct points for
'k 'equals what?

to be honest I do not know where to start

2. Nov 21, 2009

### zeion

It's a parabola and a line.

3. Nov 21, 2009

### TyErd

yeah....but how do you find k

4. Nov 21, 2009

### Staff: Mentor

At any point of intersection of the two curves, the coordinates x and y will satisfy both equations.

You have y=kx+3 and y = x^2 + 8x. Set the right sides of these equations equal and solve for x.

5. Nov 21, 2009

### TyErd

how can you solve for x when you dont know k

6. Nov 21, 2009

### Bohrok

You're not finding an exact value for x, just getting it by itself and leaving the equation in terms of k.

7. Nov 21, 2009

### Mentallic

x will be in terms of k.
Once you set the quadratic in the form $ax^2+bx+c$ with a,b,c any number or variable in terms of k, use the quadratic formula to solve for x.

Now, once you have x=... what do you need in order to have 2 distinct points? It is the same thing as thinking of what you would need to have 2 distinct roots to a quadratic equation.

8. Nov 21, 2009

### TyErd

Can anyone actually show me their working out so I can gain a more coherent understanding?

9. Nov 22, 2009

### HallsofIvy

Better if you show us your working. You learn math by doing, not by watching!

As Mark44 told you to start with, set y= kx+ 3= x^2+ 8x. Write that in the form ax^2+ bx+ c= 0 and use the quadratic formula. Yes, there will be a k in the formula.

You don't have to actually solve the equation to answer this question. Since the question asks about two different points of intersection, you want to decide what k must be so that equation has two distince solutions. A quadratic equation can have 0, 1, or 2 solutions depending on the "discriminant", b^2- 4ac.

10. Nov 22, 2009

### TyErd

11. Nov 22, 2009

### TyErd

so once rearranged it will be x^2+(8x-kx)-3=0
Use the quadratic formula and it is: (-8+k +/- sqrt k^2-16k+76 ) / 2
am i correct up to that point?

12. Nov 22, 2009

### HallsofIvy

Yes, what you have is correct.

And, as I said, you really only need the discriminant: (8- k)^2+ 12= k^2-16k+ 76. For what values of k is that positive?

13. Nov 22, 2009

### Staff: Mentor

You are solving for x, so you really should have x on one side of an equation.

x = [(-8 + k) +/- sqrt(k^2 - 16k + 76)]/2

14. Nov 22, 2009

### TyErd

Sorry HallsofIvy I am having trouble understanding what you meant by "For what values of k is that positive".

the discriminant is k^2-16k+76, what do you do with that

15. Nov 22, 2009

### Staff: Mentor

You have k^2 - 16k + 76 as the discriminant. For what values of k is the discriminant > 0? That's what HallsOfIvy asked. Do you know how to solve a quadratic inequality?

16. Nov 22, 2009

### TyErd

umm..unfortunately i don't

17. Nov 22, 2009

### Staff: Mentor

Do you know how to solve a quadratic equation?

18. Nov 22, 2009

### TyErd

19. Nov 22, 2009

### Staff: Mentor

Or however.

20. Nov 22, 2009

### TyErd

okay so i have k^2-16k+76, if i use the quadratic formula it is (16+/-sqrt (-16)^2 - 4*1*76 )/ 2, but this gives an error on my calculator

21. Nov 22, 2009

### Mentallic

Just like when you solve any quadratic, when you get the error (which means the discriminant is negative, and you can't take the square root of a negative), how does this describe the properties of the parabola?

For e.g. Take a look at the graph of $y=x^2-1$
and solve x in $x^2-1=0$

Now take a look at $y=x^2+1$
and solve x in $x^2+1=0$

can you see what is happening and understand why you get an error on the second one?
When you're solving for x, you're basically saying where does y=0.

22. Nov 22, 2009

### TyErd

yeah okay i get that, so how am i suppose to get value/s for k

23. Nov 23, 2009

### Mentallic

Well, for what values of x is $x^2+1>0$ ?

In a similar fashion, the discriminant was $(k-8)^2+12$. For what values of k is $(k-8)^2+12>0$. And the answer to this gives you the values of k where the quadratic and the line has 2 distinct roots.

24. Nov 23, 2009

### TyErd

ohk, so (k-8)^2+12>0
(k-8)^2>-12

then what would do since you cannot square root a negative

25. Nov 23, 2009

### Mentallic

No, no. I'm just asking you to use logic here.
What are the possible values of $x^2$ for all x? In other words, what is the range of $x^2$ for all values of x?
Now, similarly to that, what is the range of $(k-8)^2$ for all values of k?

What if you add 12 to this range now? So finally, for what values of k is $(k-8)^2+12>0$?