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KJ4UTS

  • one year ago

Suppose one x-intercept of a parabola...

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  1. KJ4UTS
    • one year ago
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  2. KJ4UTS
    • one year ago
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    @Operand

  3. anonymous
    • one year ago
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    Some what, though I cannot remember if you use simple form (y = ax^2 +bx +c) or standard form to solve such a question. When using standard form, you usually input everything, and then set the equation to zero, and factor to find the zeros (x-intercepts).

  4. KJ4UTS
    • one year ago
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    so this doesn't go in vertex form.

  5. anonymous
    • one year ago
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    \(y = ax^2 + bx + c\) is simple form. \(y = a(x - h)^2 + k\) is standard form. \(y = a(x - (4.7))^2 + (9.2)\) Complete the square, to get the two zeros.

  6. anonymous
    • one year ago
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    You can also make 3 equations with 3 variables. You have the general form: \[y=ax^2+bx+c\] And then you have the vertex equation: \[(T_x,T_y)=(-b/2a,-d/4a)\] And then you can make the equations: \[0=a*2^2+b*2+c\]\[4.7=-b/2*a\]\[9.2=-(b^2-4*a*c)/4*a\] Now if you solve this you get the function, and then you can set it equal to 0 and find the other x-intercept.

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