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14 June, 14:12

How would you find the zeros of the function by rewriting the function in intercept form? Ex: y=x^2-12x+32

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  1. 14 June, 14:18
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    The way i do it is factor x^2-12x+32 into (x-8) (x-4) then set each of those to zero which would look like 0=x-8 and 0=x-4. Then solve each equation and your answers would be x=8 and x=4
  2. 14 June, 14:39
    0
    x² - 12x + 32 = 0

    Δ = (-12) ² - 4 * 1 * 32 = 144 - 128 = 16

    √Δ = 4

    x₁ = (12-4) / 2 = 4

    x₂ = (12+4) / 2 = 8
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