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anonymous

  • one year ago

how would you find the x and y intercepts of y=x^2+x-2?

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  1. johnweldon1993
    • one year ago
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    Well the easiest way would to be by graphing the function, but if we cant do that we can always algebraically figure it out

  2. anonymous
    • one year ago
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    how would you solve graphically? I know the first step is to make either x or y equal 0

  3. anonymous
    • one year ago
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    ** I mean algebraically

  4. johnweldon1993
    • one year ago
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    Well you are correct that is how we do it To solve for the 'y intercept' we make x = 0 \(\large y = 0^2 + 0 - 2 \) meaning that \(\large y = ?\)

  5. anonymous
    • one year ago
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    -2

  6. johnweldon1993
    • one year ago
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    Right And now the x-intercept would be making y = 0 So \[\large 0 = x^2 + x - 2\] We can factor this or just brute force through it

  7. UsukiDoll
    • one year ago
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    to find the y-intercept we let x = 0 to find the x-intercept we let y = 0 so when x = 0, y = -2 so that would mean that our y-intercept is at (0,-2)

  8. UsukiDoll
    • one year ago
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    we can factor \[\large 0 = x^2 + x - 2 \]. and then solve for x. We would have more than one solution for our x-intercepts

  9. anonymous
    • one year ago
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    thank you both!

  10. UsukiDoll
    • one year ago
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    so we need to focus on the last term 2 and the second term 1 there's only one combination 2 x 1 = 2 (so we got the last term) we need to use subtraction to obtain one 2-1 =1 so we have \[\large 0 = (x+2)(x-1)\]

  11. UsukiDoll
    • one year ago
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    now we just split this up and solve for x in each case \[\large 0 = x-1\],\[\large 0 = x+2 \]

  12. anonymous
    • one year ago
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    so xint= -2 and 1 ?

  13. UsukiDoll
    • one year ago
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    yes our x = -2 and x = 1 so our x-intercepts are (-2,0) and (1,0) because we need y = 0 to obtain the x-intercepts.

  14. anonymous
    • one year ago
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    first put y=0 so your equation will be0=x^2+x−2 then you will make it (x-1)(x+2)=0 so it will lead that x is equal to 1 and -2

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