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Math Help - finding x and y intercepts of a parabola problem

  1. #1
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    finding x and y intercepts of a parabola problem

    I cant see where im going wrong here can somebody point it out to me please. Il show my workings.
    Trying to find the x-intercept
    y= -2(x-3)^2 +8

    so when y=o
    0= -2(x+3)^2 +8
    8= -2(x+3)^2
    -4= (x+3)^2
    -4= x^2 + 6x +9
    x^2 + 6x + 13 now i dont know whats going on, if i use the formula -b +- (square root) b^2 -4ac /2a i get square root -16 which means x is not a real number? i think im going wrong somewhere on top.

    any help appreciated my math is rusty
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by dragon555 View Post
    I cant see where im going wrong here can somebody point it out to me please. Il show my workings.
    Trying to find the x-intercept
    y= -2(x-3)^2 +8

    so when y=o
    0= -2(x+3)^2 +8
    8= -2(x+3)^2
    -4= (x+3)^2
    -4= x^2 + 6x +9
    x^2 + 6x + 13 now i dont know whats going on, if i use the formula -b +- (square root) b^2 -4ac /2a i get square root -16 which means x is not a real number? i think im going wrong somewhere on top.

    any help appreciated my math is rusty
    You don't need to expand to solve this problem and you also lost a minus sign on the 2nd line

    \displaystyle 0=-2(x+3)^2+8 \iff -8=-=2(x+3)^2 \iff 4=(x+3)^2

    Now we can just take the square root of both sides to get

    \displaystyle \pm 2 = (x+3) \iff -3 \pm 2=x

    So x=-5;x=-1
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  3. #3
    Super Member TheChaz's Avatar
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    Quote Originally Posted by dragon555 View Post
    ...which means x is not a real number? ...
    As noted, you did make a mistake, but it's also possible that there be no x-intercepts (i.e. real solutions when y = 0).
    In vertex form:
     y = a(x - h)^2 + k

    If "a" and "k" are both positive (negative), then the parabola will have no (real) x-intercepts.
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  4. #4
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    ah i see what happed on the second line with the minuses. thanks for the help guys now i can draw the correct graph lol
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