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Math Help - needing help with some problems.

  1. #1
    Junior Member mcdanielnc89's Avatar
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    needing help with some problems.

    I don't remember how to do these ones. can someone help me as much as possible? thanks!

    I have a test comeing up for finals and owul like to be able to do these without hesitation...
























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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mcdanielnc89 View Post
    I don't remember how to do these ones. can someone help me as much as possible? thanks!

    I have a test comeing up for finals and owul like to be able to do these without hesitation...
    now these look like something that needs to be graded. we can't do them for you. please tell us what you have done or thought about and we will guide you.
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  3. #3
    Junior Member mcdanielnc89's Avatar
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    i assure you these are not graded... They are example of the kinds of problems i will be doing for my final.
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by mcdanielnc89 View Post
    I don't remember how to do these ones. can someone help me as much as possible? thanks!

    I have a test comeing up for finals and owul like to be able to do these without hesitation...





    {\color{red}f(-2)=\frac{(-2)-7}{13(-2)-6}}





    {\color{red}f(x)=-\frac{1}{5}(x-0)^2+0}

    Vertex is at (0, 0) and the equation of the line of symmetry is x=0.

    Answer: C





    {\color{red}f(x)=x^2+10x+27} Now convert to vertex form

    {\color{red}f(x)=x^2+10x+25+27-25}

    {\color{red}f(x)=(x+5)^2+2}

    Vertex (-5, 2) and equation of axis of symmetry is x = -5

    Here are a few of your solutions. Others may follow.
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by mcdanielnc89 View Post
    I don't remember how to do these ones. can someone help me as much as possible? thanks!

    I have a test comeing up for finals and owul like to be able to do these without hesitation...








    {\color{red}m(5)=600(1.5)^5 \approx 4556}

    Can't see your graphs!





    Vertex (-6, 9), a>0, parabola opens upward. Answer: A



    {\color{red}x^2-2x-15=0}

    {\color{red}(x-5)(x+3)=0}

    {\color{red}x=5 \ \ or \ \ x=-3} x-intercepts

    Let x=0 in original function and find y-intercept = (0,15)





    Just substitute each function value into your original function and solve.



    Vertex is at (4, 3), parabola opens downward, graph is B. Max value = 3




    Already did one like that.

    Here's a few more. I'll let others finish. Gotta go.
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  6. #6
    Junior Member mcdanielnc89's Avatar
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    Call me stupid, but how do i do it with a - AND a constant with it?
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    Quote Originally Posted by mcdanielnc89 View Post


    Call me stupid, but how do i do it with a - AND a constant with it?
    If the coefficient of the quadratic term is negative, the parabola opens downward. That eliminates choice A as an answer.

    Set f(x)=0, and solve for x to find the zeros of the function (where the parabola crosses the x-axis)

    -3x^2+7x-4=0

    3x^2-7x+4=0

    (3x-4)(x-1)=0

    x=\frac{3}{4} \ \ or \ \ x=1

    That looks like choice B.

    The x-coordinate of the vertex is \frac{-b}{2a} in the general form ax^2+bx+c=0

    x=\frac{-7}{-6}=\frac{7}{6}

    To find the y-coordinate of the vertex, find f(7/6) in your original function f(x)=-3x^2+7x-4
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  8. #8
    Junior Member mcdanielnc89's Avatar
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    Now convert to vertex form





    Vertex (-5, 2) and equation of axis of symmetry is x = -5




    Where did the 25 come from?

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  9. #9
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    Quote Originally Posted by mcdanielnc89 View Post


    Now convert to vertex form





    Vertex (-5, 2) and equation of axis of symmetry is x = -5




    Where did the 25 come from?
    You have to take the original function f(x)=x^2+10+27 and complete the square in order to convert it to vertex form. Are you familiar with "completing the square"?

    Step 1: Take half the coefficient of x and square it. That would be 1/2 of 10, which is 5 and 5 squared = 25. That's where I get the 25. I must add that value to the function in order to make x^2+10+25 a perfect square trinomial; namely (x+5)^2

    Step2: I can't forget about the 27 I already had, and I must subtract the 25 that I added to make my perfect square trinomial. Now everything is balanced again.

    f(x)=x^2+10x+5^2 \ \ +27 \ \ - 25

    f(x)=(x+5)^2+2

    This is now in the form: f(x)=a(x-h)^2+k where (h, k) is the vertex of the parabola.
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  10. #10
    Junior Member mcdanielnc89's Avatar
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    Ok I'm terribly sry to ask agian, but i do not understand these two types of problems yet...

    I have no clue how to even find them... Anyone explain?



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  11. #11
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    Quote Originally Posted by mcdanielnc89 View Post
    Ok I'm terribly sry to ask agian, but i do not understand these two types of problems yet...

    I have no clue how to even find them... Anyone explain?




    Ok, here goes. You can find the vertex of a quadratic function (parabola) two ways. You can convert the function from standard form f(x)=ax^2+bx+c to vertex form f(x)=a(x-h)^2+k, in which (h, k) is the vertex.

    The other way is to find the x-coordinate of the vertex and then use that in the function f(x) to find the y-coordinate of the vertex. The x-coordinate of the vertex is \frac{-b}{2a}.

    \frac{-b}{2a}=-\frac{-5}{2}=\frac{5}{2}. This is your x-coordinate of the vertex.

    f(x)=x^2-5x

    f\left(\frac{5}{2}\right)=\left(\frac{5}{2}\right)  ^2-5\left(\frac{5}{2}\right)=\frac{25}{4}-\frac{25}{2}=\frac{25}{4}-\frac{50}{4}=-\frac{25}{4}

    So the vertex is \left(\frac{5}{2}, -\frac{25}{4}\right)

    The parabola opens upward since the coefficient of x^2 is positive.

    The graph looks to be answer D.

    The axis of symmetry is the vertical line that splits the parabola into 2 symmetric halves through the x-coordinate \frac{5}{2}.

    Equation of axis of symmetry is : x=\frac{5}{2}



    See if you can do the second one. I'll help you, or someone will help you if you get stuck.
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  12. #12
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    Quote Originally Posted by mcdanielnc89 View Post
    Ok I'm terribly sry to ask agian, but i do not understand these two types of problems yet...

    I have no clue how to even find them... Anyone explain?



    These look strangely familiar.

    First, the x-coordinate of the vertex can be found using the formula x=-\frac{b}{2a} in the function f(x)=ax^2+bx+c

    So, we have x=-\frac{-5}{2}=\frac{5}{2}

    This also helps with the axis of symmetry since it is a vertical line that passes through the x-coordinate of the vertex (which we just found).

    Axis of symmetry: x=\frac{5}{2}

    Now for the y-coordinate of the vertex to complete the ordered pair.
    find f(\frac{2}{5})\ \ in \ \ f(x)=x^2-5x. Pair that up with the x-coordinate and you have the coordinates of the vertex.

    As far as the graph is concerned, the parabola opens upward since the coefficient of the quadratic term is positive. After you determine the y-coordinate of the vertex as outlined above, you could conclude that the graph is answer D.


    Same as before. The x-coordinate of the vertex is x=\frac{-b}{2a}. Plug that answer into your original function to solve for
    f(x) or y. Now you have the y-coordinate of the vertex.

    The axis of symmetry is x=the x-coordinate of the vertex.

    The parabola opens downward since the coefficient of the quadratic term is negative.
    Here you go!
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