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 $(k-8)^2+12>0$?