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Math Help - Parabolas

  1. #1
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    Parabolas

    How do I figure out problems similar to the following by graphing y = 2x2 + 3?
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  2. #2
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    What are you trying to find..Also your equation  y = 2x2 + 3 , does it mean y=4+3 or y=4x+3 or...you just mistype and its suppose to be y=2x+3?

    Edit
    Never mind, I assume you mean  y=2x^2+3 right? But the question still exists, what are you trying to find?
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  3. #3
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    I mean how do I graph it as a parabola?
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  4. #4
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    Find the y intercept, which means setting x=0, and the y intercept in this case is (0,3)
    Find the x intercepts using the quadratic equation or factoring- There is no x-intercepts in that problem
    Find the axis of symmetry, which is  x=-b/(2a) to find the min or the max otherwise known as vertex, in this case it would be the same as the y-intercept
    Look at the sign to see if it's concave up or down
    I think thats basically it to graph simple parabolas like the example you gave.
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  5. #5
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    I'm sorry but I am still a bit confused... Is there a simpler way?
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  6. #6
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    Easier way...plug in x values...or use the calculator
    I assume you know what vertex, and setting x=0 and y=0 does since this is in the pre-calc section
    The basic equation of a parabola is y= ax^2+bx+c
    So to find the vertex of the equation you gave y=2x^2+3 , you would do x=-b/2a, and in this cause there is no b, which means it is 0 so the x value of the vertex would be x=0/4=0 and plug in 0 for x in the equation you would get the vertex at (0,3).

    Now that you have the vertex it's time to find the x intercept. To find the x-intercept you would either use the quadratic equation or factoring it if possible. The equation equation is:
    Now solve for x. Whatever you get for x should the x intercepts..which I assume you know what it is. I don't think there is any x-intercepts in this case.
    Now for the y intercept just set x=0 in the equation. Because that means when x=0, y=?. In this case when you set x=0, y will be 3 right?
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  7. #7
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    I understand everything in this part...
    Easier way...plug in x values...or use the calculator
    I assume you know what vertex, and setting x=0 and y=0 does since this is in the pre-calc section
    The basic equation of a parabola is
    So to find the vertex of the equation you gave , you would do x=-b/2a, and in this cause there is no b, which means it is 0 so the x value of the vertex would be x=0/4=0
    But I don't understand how you got the vertex....
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  8. #8
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    -b/2a is the x value of the vertex because it is also the axis of symmetry for the parabola. To get the vertex, you plug in the x and you will get the y-value so the vertex would be (-b/2a, f(-b/2a)) with f(x) = y. If you are looking for where did I get -b/2a or how to derivative it, read this webpage
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  9. #9
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    Is it any different to graph a linear equation such as y=2x^2+4? I don't know if this is a linear equation....
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  10. #10
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    That is not a linear equation. It is a quadratic equation and no, it is not any different to solve   y=2x^2+4
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  11. #11
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    Can you please give an example of a linear equation and what makes it linear? thanks
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  12. #12
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    Anything in the form of y = mx + b where m is the slope and b is the y-intercept.

    It is linear because when you graph it you do not get a curve. You get a straight line.

    And to do this really easily just plug in x-values, from that you can get the y-values and then plot. Then do a simple connect the dots. You should get a pretty skinny curve with the vertex at (0,3) You can find the vertex as linnus said by doing \left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right) You can use this only if it is in the form ax^{2}+bx+c
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