E needs algebra help again

[klmno]

And I'm letting him explain the question:


I need to know how to get the x and y intercept and the vertex of f(k)=k^2-14k-51. It just dosn't add up to me because i know (k-7)^2= k^2-14k+49. But, I'm not sure where the -51 came from. Also my teacher gave me the ansers for this one but i still dont understand how to get it: the y intercept=(0,-51) and the x intercepts= (17,0) and (-3,0) I'm sure if I could understand how to get this one I will be able to figure out my other 19 problems. Thanx hope you-all can help me. -E

 

[alongfortheride]

http://www.khanacademy.org/

 

[TeDo]

To find the y intercept you need to put a 0 (zero) in for x and solve. In this case y= 0^2 - 14(0)-51 or y = -51. To find the x intercepts, you have to make the y a zero and solve. To do that you need to find all the factors of 51 (1,3,17,51). Since the two signs are both negative, the larger factor must be negative and the smaller one is positive because when they are multiplied, they need to be negative and when added, then also need to be negative. Here's what I got.

0=k^2 - 14k - 51
0= (k-17)(k+3)

Then you set each one equal to zero and solve like this:

0=k-17 and 0=k+3
17=k and -3=k

The y intercept is (0, -51) and the x intercepts are (17,0) and (-3, 0).

Does that make sense?? I really hate doing the work FOR kids but since you had the answers but didn't understand the process, the only way to explain the process is to show it.

 

[klmno]

Hey im E yes thankyou that made a lot better sense then the way my teacher did it. Now, I can do majority of the homework with this information but could you try to explain how to get the vertex. The reason I don't know is because I had a subsitute teacher and my regular teacher left this with her to give to us as homework but it's a tad bit different than what we are used too. Thanx again for the help.

 

[klmno]

Thanks Ladies! E did the majority of it annd will have to live with the rest. I don't know if it's the teacher or him but for some reason he can get it when it's esplained a different way.

 

[Kathy813]

Good job, Tedo. I couldn't have done it better myself.

To find the vertex, you use the equation x = -b/2a (or in this case: k = -b/2a).

f(k) = k^2-14k-51

a=1; b = -14; k = -51

So for the vertex:

k = -(-14)/2(1)
k=14/2
k=7

Then subsitute the k value (7) into the equation for k and that will give you the y-value of the vertex.

f(7) = (7)^2 - 14(7) - 51
f(7) = -100

Therefore, the vertex is at (7, -100).

There is more than one way to do this. You could also use completing the square to change the equation into vertex form and find the vertex that way. I don't know if he is allowed to use a graphing calculator but you could also find the vertex by using the second trace function on the calculator. Still another method if you know the x-intercepts (which you do in this case) is to take the two x-values, add them together and divide by two since the vertex will fall halfway between them. Then do the second step above.

Is your head spinning yet? LOL

I love math!!

 

[klmno]

Apparently so, Kathy. I just got home and E said he went to class with a little over half his problems done, or so he thought. They'd had a substitute Friday and his regular teacher was there today. He said no one had understood correctly and everyone had trouble with the homework so this regular teacher explained it, it made sense to E, and she gave the whole class one more day. He said she wanted it done in a different method than what he'd done.

I really appreciate the help here- I wouldn't have asked a second time but I could tell he was trying and made every effort himself to find it in the textbook and reason it out. The teacher doesn't use the textbook though.

 

[Star*]

I am absolutely amazed at you girls......and you too E!!!!!!

I have no earthly idea what either of you just said -OR What it 's used for but WOW!!!!!!!!

Way To Go E!!!!! YOU ROCK MAN!