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Due Tomorrow Please Help Now

  1. Nov 17, 2004 #1
    Due Tomorrow Please Help Now!!

    #2 Tell Whether The Quadratic Has A Maximum Value Or A Minimum Value. The Find The Value. Round To The Nearest Tenth. :confused:

    F(x)= -x² -6x - 7


    #3 Let y = | x+a | +b, Where a ‡ 0 and b ‡ 0 . Explain how the values of a and b affect the graph of the functions as compared to the graph of y= |x|.



    I NEED HELP FAST HURRY! :cry: THANK YOU!
     
    Last edited: Nov 17, 2004
  2. jcsd
  3. Nov 17, 2004 #2

    James R

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    Hints:

    #1 Solve. Round Decimal Answer To The Nearest Tenth.

    x + y = 6 -(1)
    y + z = 4 -(2)
    x + z = 2 -(3)

    Take equation (1). Rearrage to get y in terms of x.
    Substitute that into equation (2) to get a new equation for x and z - equation (2b)
    Eliminate either x or z from equations (2b) and (3), solving simultaneously.
    Back-substitute to get the other two solutions.

    #2 Tell Whether The Quadratic Has A Maximum Value Or A Minimum Value. The Find The Value. Round To The Nearest Tenth.

    F(x)= -x² -6x - 7

    This is a parabola. Which way round does it go? Peak at the top, or the bottom? How can you tell?

    To find the maximum or minimum, you can either differentiate (if you know calculus), or complete the square (if you don't).

    #3 Let y = | x+a | +b, Where a ‡ 0 and b ‡ 0 . Explain how the values of a and b affect the graph of the functions as compared to the graph of y= |x|.

    How do the following graphs differ:

    y = sin x
    y = sin x + 6

    How about these ones:

    y = sin x
    y = sin(x+6)

    Apply that to your problem.

    If you need more help, let us know.
     
  4. Nov 17, 2004 #3
    I Need More Help. I DONT GET IT.
     
  5. Nov 17, 2004 #4
    Well, lets start with the first one:

    x + y = 6
    y + z = 4
    x + z = 2

    You have three equations and three unknowns. Pretty simple here.

    Lets write the first equation in terms of y.

    y = 6 - x

    (*Remember when you move a term to the other side of the equals sign, its sign changes*)

    Now, we know what y is. So we can take that and substitude (6-x) wherever there is 'y' term.

    Take a look at the second equation:

    y + z = 4

    We now know what y is, so we can turn this into a new equation.

    (6 - x) + z = 4
    6 - x + z = 4
    z - x = -2

    Now lets take a look at the third equation

    x + z = 2

    Lets rearrange this bad-boy in terms of z

    z = 2 - x

    (*Isn't this fun?*)

    Knowing that
    z - x = -2
    We can now write
    2 - x - x = -2
    -2x = -4

    I hope you can take it from there.
     
  6. Nov 17, 2004 #5
    thank you, how about the 2nd and the 3rd ones?
     
  7. Nov 17, 2004 #6
    need answers
     
  8. Nov 17, 2004 #7
    Well, for the second question lets look at your equation.

    F(x)= -x² -6x - 7

    Its in the form ax^2+bx+c which means that you are given a parabola


    This is what a f(x) = x^2 looks like

    [​IMG]

    Now what we have here is a little different. Our parabola has a negative sign infront of it. This means that it will be upside down, or opening downwards.

    From that you should be able to figure out if it will have a MAX or a MIN.

    To find the MAX/MIN value just set dy/dx(f(x)) = 0

    Solve for x, plug into f(x) and you'll get your y value.
     
  9. Nov 17, 2004 #8
    I'll put in just as much effort as you do, chikawakajones.

    #2: We hav

    That's about it.

    --J
     
  10. Nov 17, 2004 #9
    how do u solve for x?
     
  11. Nov 18, 2004 #10
    haha :rofl:
     
  12. Nov 18, 2004 #11
    just help me...........
     
  13. Nov 18, 2004 #12
  14. Nov 18, 2004 #13
    Well, you have two options. If you know calculus you can take the derivative of f(x), or you can complete the square.

    Either way is really easy. And you'll get the same value for x.

    I'm going to assume right off the bat that you don't know calculus. So, I'll explain how to complete the square

    f(x) = -x^2-6x-7


    For the general equation:

    x^2 + bx + c

    Take half of the middle term (b) and add its square to the last term (c).

    (x-b/2)^2 + c + b^2

    set

    x-b/2 = 0

    solve for x

    plug x into F(x)

    (x, F(x) ) are your points
     
    Last edited: Nov 18, 2004
  15. Nov 18, 2004 #14
    im doing other problems right now.....that is why i need as much help i could get
     
  16. Nov 18, 2004 #15
    for the x value, i got -22. is this right?
     
  17. Nov 18, 2004 #16
    Err--no. Not quite sure how you got that.

    Lets look at this again, and break it down.

    We have

    f(x)= -(x^2) -6x -7

    Lets break this bad boy down.

    Term A is going to be [tex]-(x^2)[/tex]

    Term B is going to be [tex]-6x[/tex]

    Term C is going to be [tex]-7[/tex]

    Lets take out the x's

    b = 6

    c = -7

    Now the general formula is

    [tex](x-\frac{b}{2})^2 + (\frac{b}{2})^2 + c[/tex]


    Now take your [tex](x-\frac{b}{2})^2[/tex] part

    and set [tex]x-\frac{b}{2} = 0[/tex]

    solve for x

    plug into your oringal function to get y

    Or you know what in this case just take b (READ: 6) divide it by two (READ: 6/2) and then set x - b (READ: -x - 6/2) to 0 (READ: x + 6/2 = 0)

    Then plug that value (READ: -6/2) into f(x) to get your Y value.
     
    Last edited: Nov 18, 2004
  18. Nov 18, 2004 #17

    James R

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    If x=-22 is supposed to be the x value of the maximum point, the answer is "no - that's not right."

    Please post your working so we can tell you where you went wrong.
     
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