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Math Help - Finding x-intercepts and vertexes of quadratics.

  1. #1
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    Finding x-intercepts and vertexes of quadratics.

    ok, so i'm having a rough time trying to get an answer, mostly because i don't have confidence on my answers. (i already answered question #1)

    2) m^2 = 9m-1
    for this, i rearranged it to become m^2-9m+1=0 (in accordance w/the standard form i think). so my vertex is (4.5, 61.75) and the x-intercepts i got were {0.1125, 8.8875}. my graph is a parabola pointing downward...
    3) -25=x^2+10x
    for this, i rearranged to get y=x^2+10x+25. i got a vertex of (-5,0). i'm stuck here because i'm not even sure what i did in #2 was correct.
    * edit * i tried to get x-intercepts and i got two -5s. i really don't think my x-intercepts are right.
    thanks in advance.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by trancefanatic View Post
    ok, so i'm having a rough time trying to get an answer, mostly because i don't have confidence on my answers. (i already answered question #1)

    2) m^2 = 9m-1
    for this, i rearranged it to become m^2-9m+1=0 (in accordance w/the standard form i think). so my vertex is (4.5, 61.75) and the x-intercepts i got were {0.1125, 8.8875}. my graph is a parabola pointing downward...
    intercepts are right, vertex is wrong. i think you miscalculated when you pluged in the 4.5
    3) -25=x^2+10x

    for this, i rearranged to get y=x^2+10x+25. i got a vertex of (-5,0). i'm stuck here because i'm not even sure what i did in #2 was correct.
    * edit * i tried to get x-intercepts and i got two -5s. i really don't think my x-intercepts are right.
    thanks in advance.
    you are correct. the vertex happens to be the only x-intercept. the vertex touches the x-axis, so the x-intercept is one point, namely the vertex. trust yourself more
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    intercepts are right, vertex is wrong. i think you miscalculated when you pluged in the 4.5
    oh yeah! ok, so i recalculated and got (4.5,19.25)

    Quote Originally Posted by Jhevon View Post
    you are correct. the vertex happens to be the only x-intercept. the vertex touches the x-axis, so the x-intercept is one point, namely the vertex. trust yourself more
    so... would there be a graph at all?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by trancefanatic View Post
    oh yeah! ok, so i recalculated and got (4.5,19.25)
    it should be -19.25 i believe.


    so... would there be a graph at all?
    why would you think that? why shouldn't there be a graph? the graph is below
    Attached Thumbnails Attached Thumbnails Finding x-intercepts and vertexes of quadratics.-graph3.jpg  
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    it should be -19.25 i believe.


    why would you think that? why shouldn't there be a graph? the graph is below
    i thought that because i thought everything had to have coordinates, lol.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by trancefanatic View Post
    i thought that because i thought everything had to have coordinates, lol.
    umm, every point on the graph does have a coordinate.

    don't let the fact that the parabola has one x-intercept mess you up, there are some with no x-intercept, it just means the parabola always returns values that are positive, but it always returns values, since parabolas are defined for all x.
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