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KJ4UTS
 one year ago
Suppose one xintercept of a parabola...
KJ4UTS
 one year ago
Suppose one xintercept of a parabola...

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Some 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 (xintercepts).

KJ4UTS
 one year ago
Best ResponseYou've already chosen the best response.0so this doesn't go in vertex form.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(y = ax^2 + bx + c\) 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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: \[0=a*2^2+b*2+c\]\[4.7=b/2*a\]\[9.2=(b^24*a*c)/4*a\] Now if you solve this you get the function, and then you can set it equal to 0 and find the other xintercept.
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