Find k so y = -4x^2+kx-1 has integer for max value

Markus

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Okay thanks so much for taking the time to read this.

Here is the questions that i cannot do:

1. If y=-4x²+kx-1, determine the value(s) for k which the maximum value of a function is an integer. Explain your reasoning.

2. The graph of the function f(x) = x²-kx+k+8 touches the x-axis at one point. What is the value of k?

I need this asap! Thanks sooo much!
 
1. Complete the square as follows:
\(\displaystyle y= -4\left(x^{2} - \frac{k}{4}x\right) - 1\)
\(\displaystyle y = -4\left(x^{2} - \frac{k}{4}x + \frac{k}{64}\right) - 1 + 4\left(\frac{k}{64}\right)\)
\(\displaystyle y = -4\left(x - \frac{k}{8}\right)^{2} + \left(\frac{k^{2}}{16} - 1\right)\)
Now, we know for the parabola in the form y = A(x + B)² + C we know that the sign of A determines whether the vertex is its minimum or maximum point, B determines the horizontal shift, and C represents the VERTICAL shift which is what we're looking for. So, for what values of C would yield an integer value?

2. The determinant (i.e. the part of the quadratic formula) determines how many roots (i.e. how many times it touches the x-axis) a parabola has. If a parabola has:
2 roots, then \(\displaystyle b^{2} - 4ac > 0\)
1 root, then \(\displaystyle b^{2} - 4ac = 0\)
no roots, then \(\displaystyle b^{2} - 4ac < 0\)

Can you solve it now?
 
I still dont understand the first one. I get that we have it in the form of a porabola but what do i do now? Would i have -4(x-k²/64) + (-10k) ?? Or am i way off? Please post your answers too both questions. It seems so familiar! By the way thanks soo much, even if you stop now you have helped me a lot!
 
1. \(\displaystyle y = -4\left(x - \frac{k}{8}\right)^{2} + \left(\frac{k}{16} - 1\right)\)
Think of C as \(\displaystyle \left(\frac{k}{16} - 1\right)\). Now, with what you know about fractions, what values of k would give you an integer value? For example, if k = 8 would it work? How about k = 16? k = 32? ... See where I'm going?

Anyway, I have to go now. Hopefully you'll get it! And if not, maybe someone else may stop by and help out! Good luck.
 
Before you go please just post your answer to it. I can study it and find out how you got it. I really need this!!
 
I fixed my first post. Read it carefully and see how the examples of k apply to the problem for the first question. For the second question, if you're looking for one root, solve for the determinant equation I gave you. \(\displaystyle b^{2} - 4ac\) = 0 where a,b, and c are the coefficients of your equation (review your quadratic formula). Anyway I'm running late, good luck!
 
1) I think that i have the wrong understanding on Integer. Is it any full number? or a number that does not go into decimals? If so than i guess that the first number that makes the integer is 16. but that result is 0 so would k is 16?

2) This asks for the value of k. I dont understand were to go on from there. I know the equation but the thing is that it has more than the regular a b and c. If it was something like x² + 4x + 5 than i would be able to understand it and plug it all in but for this one there are 4 parts to it. I just dont get that part. Can you start me of on the second one please.
 
Markus said:
1) I think that i have the wrong understanding on Integer.
To learn what integers are (if this wasn't covered in your book or in class), please try search-engine results for definitions, examples, and lessons. (If, by "full number", you mean "whole numbers, and their negatives", you are on the right track.)

Markus said:
2) This asks for the value of k....for this one there are 4 parts to it.
I'm sorry, but I'm only seeing the one question for exercise for (2). If there are "parts" (a) through (d), they don't appear to have yet been posted...? :?:

In any case, this exercise tells you that x<sup>2</sup> - kx + (k + 8) has only one zero, which you know (from your lessons and homework) means that you have a particular sort of solution when you apply the Quadratic Formula to a = 1, b = -k, c = k + 8. So plug a, b, and c into the Formula you've memorized, and set the appropriate bit equal to zero. Then solve the equation for the value of k. :idea:

Markus said:
Before you go please just post your answer to it. I can study it and find out how you got it.
Most legitimate tutors don't "do" students' work for them, or give out the solutions. If you haven't been able to understand from the book, the class sessions, the set-up, and the explanation, then I'm afraid another worked example (the complete solution to your homework) wouldn't help your understanding. Sorry! :oops:

Please reply showing what you have tried, and how far you have been able to get. Once the tutors can "see" where you're going wrong, we can provide specific help that will get you moving again toward the solutions. Thank you! :D

Eliz.
 
You were given a hint for the second problem.

If the "discriminant" (the expression under the radical sign in the quadratic formula) is equal to 0, the graph of the quadratic will have its vertex on the x-axis, and will have just one solution.

If
b<SUP>2</SUP> - 4ac = 0, then there's just one solution.

Your equation is

f(x) = x<SUP>2</SUP> - kx + k + 8

"a" is the coefficient of the x<SUP>2</SUP>, so that's 1.
"b" is the coefficient of the x term, so that's -k.
"c" is the constant terms, and that's (k + 8)

(-k)<SUP>2</SUP> - 4*1*(k + 8) = 0

Now....solve for k.
 
2) When i start to solve that this is what i get.

Since b²-4ac the equation is (-k)²-4(1)(k+8) and than i get k-4k-32=0 and this is were i get stuck. Where do i go from here? I cant factor it since its not in the right form and i cant really do anything.

edit: Wait, would i switch everything so its k= 4k+32...but if i do it still wont make sense because i cant add 4k and 32. What if i divide the right side by 4 and than switch the k to the left making it 2k and than just divide both sides by 2. But i think that is wrong since you cant just divide the one side right?
 
2) ohh thats great my answer did not have k^2 i simply had k and thats why i did not think it was able to be factored. Okay so from now on i just factor it right?
 
2) Actually there should be a k² in your expression. Recalling that the determinant should equal to zero: \(\displaystyle b^{2} - 4ac\). We have that b = -k, a = 1, c = k + 8. So, it should look something like this:

\(\displaystyle b^{2} - 4ac = 0\)

\(\displaystyle (-k)^{2} - 4(1)(k + 8) = 0\)

And from there you should see that you have an expression with the k², which as mentioned earlier, means that you have to factor.

1) Yes, you can think of an integer as a whole number that includes the negatives. So could there possibly be more than one value for k in question 1?
 
1) Well im not sure. So far the only one that i have is 16. Is that right so far? If not please give me a hint as to what i did wronge.

2) Also when i factored it i got. (k-8)(k+4) is that the answer?
 
1) Looking at C again: \(\displaystyle \frac{k}{16} - 1\)

You are correct that k = 16 would work because \(\displaystyle \frac{16}{16} - 1 = 0\) which is an integer. How about k = 32? \(\displaystyle \frac{32}{16} - 1 = 1\) which also is an integer. How about k = 64? 80? 1600? Really, I'm just asking what do these numbers have in common.

2) Yes, you factored it correctly. So what values of k would make (k-8)(k+4) equal to zero?
 
o_O said:
1) ...I'm just asking what do these numbers have in common.
They are all multiples of 16 for example 16x1 16x2 etc. Okay so for the first question what would my answer be? I know that they are all multiples of 16 so since the question is "Determine the value(s) for k which the maximum value of a function is an integer" would I put "16, 32, 48, 64, etc..?"

o_O said:
2) Yes, you factored it correctly. So what values of k would make (k-8)(k+4) equal to zero?
8 and -4 would make (k-8)(k+4) equal to zero right? But even if that is the question asks "What is the value of k?" and those are two numbers.
 
look at the original question again ...

If y=-4x²+kx-1, determine the value(s) for k which the maximum value of a function is an integer. Explain your reasoning.
the max value of a quadratic function that opens down (as this one does) occurs at the vertex of the parabola, or where x = -b/(2a) = -k/(2*-4) = k/8

evaluating the quadratic at x = k/8 yields ...

y(k/8) = -4(k/8)² + k(k/8) - 1 = -k²/16 + k²/8 - 1 = k²/16 - 1 = (k/4)² - 1

now, look at the last expression ... if the max value of y is an integer, then k has to be a multiple of 4.

I apologize, but it seems to me that you were getting farther and farther away from what the original question is asking.
 
skeeter said:
the max value of a quadratic function that opens down (as this one does) occurs at the vertex of the parabola, or where x = -b/(2a) = -k/(2*-4) = k/8

evaluating the quadratic at x = k/8 yields ...

y(k/8) = -4(k/8)² + k(k/8) - 1 = -k²/16 + k²/8 - 1 = k²/16 - 1 = (k/4)² - 1
You just lost me... Im soo confused! Why is there a k/8 on the left side? The equation is not yx...
 
Terribly sorry. Forgot to square the k for "C":

\(\displaystyle \left(\frac{k^{2}}{16} - 1\right) = \left(\left(\frac{k}{4}\right)^{2} - 1 \right)\)

So skeeter is right on the multiple of 4. I mistakenly was thinking a multiple of 16 but that was because I forgot to square k when completing the square.

What skeeter was doing was basically another approach to finding the "C" value I had. Because of how the parabola is shaped, we know that the maximum value of the function will occur at its vertex. So, he suggested finding the x-coordinate by a standard 'formula' and then plugged it into the original function to yield "C" - the y-coordinate / vertical shift - however you want to call it.
 
eather way it does not matter. What would i put down for the answer? I know its a multiple of 4 but what does that matter? and for the second one. Now that i have (k-8)(k+4) and i know that to make them zero it is 8 and -4 what do i do now?
 
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