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Math Help - Graph Problem

  1. #1
    Member GAdams's Avatar
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    Graph Problem

    Could somone please provide some pointers on how to tackle these sorts of questions. I find them a bit scary.
    Attached Thumbnails Attached Thumbnails Graph Problem-graph.jpg   Graph Problem-graph-2.jpg  
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    Could somone please provide some pointers on how to tackle these sorts of questions. I find them a bit scary.
    For the first question.

    the graph given is y = 2x^2 + x - 4

    the x-intercepts give the x-values for which y = 0

    so for this graph, the x-intercepts are the x-values such that:

    2x^2 + x - 4 = 0

    or in other words, where 2x^2 + x = 4

    so the solutions to the first question is simply the x-intercepts, which you can tell by looking on the graph


    For the second question:

    we want the solutions to 2x^2 - x - 5 = 0

    this is where 2x^2 + x - 4 = 2x + 1

    So draw the line y = 2x + 1, the x-values of the points of intersection are your solutions
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  3. #3
    Member GAdams's Avatar
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    Quote Originally Posted by Jhevon View Post
    For the first question.

    the graph given is y = 2x^2 + x - 4

    the x-intercepts give the x-values for which y = 0

    so for this graph, the x-intercepts are the x-values such that:

    2x^2 + x - 4 = 0

    or in other words, where 2x^2 + x = 4

    so the solutions to the first question is simply the x-intercepts, which you can tell by looking on the graph


    For the second question:

    we want the solutions to 2x^2 - x - 5 = 0

    this is where 2x^2 + x - 4 = 2x + 1

    So draw the line y = 2x + 1, the x-values of the points of intersection are your solutions


    I can't see where the 2x + 1 from 2x^2 + x - 4 = 2x + 1 came from. Thanks.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    I can't see where the 2x + 1 from 2x^2 + x - 4 = 2x + 1 came from. Thanks.
    we find solutions from graphs based on intersections. therefore we need to look for those desired intersections, or sometimes create them.

    in the following problem we were asked to draw a line to find solutions to the equation 2x^2 - x - 5 = 0, but i know nothing about that equation from the graph, all i know about from the graph is the equation y = 2x^2 + x - 4, so i have to somehow create this graph from the new graph given so i can get the required intersections.

    so we begin with the new graph given:

    2x^2 - x - 5 = 0 (remember, my objective is to extrapolate 2x^2 + x - 4 = \mbox { something } from this graph

    since i need a +x i have to add 2 x's to -x, but since this is an equation, i have to add it to both sides, so i get:

    2x^2 + x - 5 = 2x

    now i need to get a -4 instead of a -5, so i add 1 to both sides, we get:

    2x^2 + x - 4 = 2x + 1

    I changed the graph of 2x^2 - x - 5 = 0 to get this, so my solutions to the new equation are exactly the same solutions to the one asked. since this new equation has the graph that i have originally, i simply need to find the intersection points between my old graph and the line 2x + 1

    Quote Originally Posted by Jhevon View Post
    For the first question.

    the graph given is y = 2x^2 + x - 4

    the x-intercepts give the x-values for which y = 0

    so for this graph, the x-intercepts are the x-values such that:

    2x^2 + x - 4 = 0

    or in other words, where 2x^2 + x = 4

    so the solutions to the first question is simply the x-intercepts, which you can tell by looking on the graph


    For the second question:

    we want the solutions to 2x^2 - x - 5 = 0

    this is where 2x^2 + x - 4 = 2x + 1

    So draw the line y = 2x + 1, the x-values of the points of intersection are your solutions
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