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Math Help - Quadratics Question

  1. #1
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    Quadratics Question

    I was wondering if someone would be kind enough to walk me through how to do this problem:

    Describe when the parabola y=-3x(power of 2) + 15x + 42 crosses the x-axis. Confirm your answer by graphing the equation.

    Thank you for the help guys
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  2. #2
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    Note that when a point lies on the x-axis, it has a y-coordinate of 0. So in order to find your roots, we set your quadratic equation equal to 0, i.e.:
    0 = -3x^2 + 15x + 42

    First, divide -3 from both sides since we're going to have to factor and it's much easier to deal with no coefficients in front of the x^{2}:
    0 = x^{2} - 5x - 14

    Now factor the quadratic to get the form: 0 = (x - a)(x-b) and solve for x by setting x - a = 0 and x - b = 0.

    This works because when you multiply 2 numbers to get 0, either one or the other or both are equal to 0. So, setting both terms (x-a) and (x-b) equal to 0 will lead you to your x-values that correspond to the y-intercept.
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    Thank you!

    So having it in the factored form gives me enough information to graph it as well?

    Well, actually.. I don't quite understand how one would graph this out from just the factored form.
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    Here's a smart way to think about it. You have your two x-intercepts right? This is enough information to find the x-value of your vertex since the quadratic (parabola) is symmetric about it. Your x-value for the vertex should be dead centre between your two x-intercepts.

    Then, plug this x-value back into your original quadratic to find its value. Now, you have your vertex and x-intercepts which is enough to graph it. (You may also want to find the y-intercepts by simply setting x = 0 in your original quadratic).
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  5. #5
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    Quote Originally Posted by o_O View Post
    Here's a smart way to think about it. You have your two x-intercepts right? This is enough information to find the x-value of your vertex since the quadratic (parabola) is symmetric about it. Your x-value for the vertex should be dead centre between your two x-intercepts.

    Then, plug this x-value back into your original quadratic to find its value. Now, you have your vertex and x-intercepts which is enough to graph it. (You may also want to find the y-intercepts by simply setting x = 0 in your original quadratic).
    Well thank you again o_O it makes sense now!

    I appreciate your time.

    How do you figure out how high along the y-axis the vertex is though?

    Sorry to bother you again.
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  6. #6
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    Quote Originally Posted by bliks View Post
    Thank you!

    So having it in the factored form gives me enough information to graph it as well?

    Well, actually.. I don't quite understand how one would graph this out from just the factored form.
    Edit: Too slow, but here it is anyway.

    Plot the x-intercepts found after you factor the quadratic.

    Plot the y-intercept. In y=ax^2+bx+c, the y-intercept is "c".

    Convert y=ax^2+bx+c to vertex form and find the turnaround point. The parabola opens downward since the coefficient of  x^2 is negative.

    If you need more points to plot, set up a table using

    x and f(x). Assign arbitrary values for x and solve for f(x).
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  7. #7
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    Quote Originally Posted by masters View Post
    Edit: Too slow, but here it is anyway.

    Plot the x-intercepts found after you factor the quadratic.

    Plot the y-intercept. In y=ax^2+bx+c, the y-intercept is "c".

    Convert y=ax^2+bx+c to vertex form and find the turnaround point. The parabola opens downward since the coefficient of  x^2 is negative.

    If you need more points to plot, set up a table using

    x and f(x). Assign arbitrary values for x and solve for f(x).
    That makes sense thank you so much for your time.
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  8. #8
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    Quote Originally Posted by bliks View Post
    Well thank you again o_O it makes sense now!

    I appreciate your time.

    How do you figure out how high along the y-axis the vertex is though?

    Sorry to bother you again.
    Don't be sorry

    Remember, anything along the y-axis has a x-coordinate = 0. So, set x = 0 in your original quadratic equation and solve for y.

    Edit: Ah too slow too!
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  9. #9
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    Quote Originally Posted by masters View Post
    Edit: Too slow, but here it is anyway.

    Plot the x-intercepts found after you factor the quadratic.

    Plot the y-intercept. In y=ax^2+bx+c, the y-intercept is "c".

    Convert y=ax^2+bx+c to vertex form and find the turnaround point. The parabola opens downward since the coefficient of  x^2 is negative.

    If you need more points to plot, set up a table using

    x and f(x). Assign arbitrary values for x and solve for f(x).
    Quote Originally Posted by o_O View Post
    Don't be sorry

    Remember, anything along the y-axis has a x-coordinate = 0. So, set x = 0 in your original quadratic equation and solve for y.

    Edit: Ah too slow too!

    Again thank you!

    So just to be sure... Y= 42 right? and the factored form comes out to be (x+2)(x-7), meaning that the zeros are -7 and 2 and the axis of symmetry would be -1?
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  10. #10
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    Not quite.

    (x +2)(x-7) = 0

    Set each term equal to 0:
    x+2 = 0 \: \: \Rightarrow \: \: x = -2
    x - 7 = 0 \: \: \Rightarrow \: \: x = 7

    You had your signs switched when solving for x.
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