Question 206260
your x intercepts are -1 and 3 and this is a quadratic equation.
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this means that (x+1)*(x-3) = 0 because that is the quadratic equation that would yield these 2 x intercepts.
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multiplying them out we get {{{x^2 - 2x - 3 = 0}}}
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that's our equation.
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a limitation is that the range of our function can't be greater than 4.
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Since y represents the range of our function, this means y has to be less than or equal to 4.
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since our equation is  y = f(x) = {{{x^2 - 2x - 3}}}, this means that {{{x^2 -2x - 3 = 4}}} would represent the maximum value that y could be.
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in order to restrict the range, we have to restrict the domain because the range is dependent on the domain.
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we need to find the x values for when y = 4
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if we subtract 4 from both sides of the equation of {{{x^2 -2x - 3 = 4}}} we get {{{x^2 -2x - 7 = 0}}}
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if we find the roots of this equation, we should be able to find the x values for when y = 4.
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solving the equation of {{{x^2 -2x - 7 = 0}}} using the quadratic formula of {{{(-b +- sqrt(b^2-4ac))/(2a)}}} yields the following answers:
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x = 3.828427125 or x = -1.828427125
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we got this in the following manner:
a = 1
b = -2
c = -7
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2a = 2
-b = 2
{{{sqrt(b^2 - 4ac)}}} = {{{sqrt(4-(4*1*(-7)))}}} = {{{sqrt(4+28)}}} = {{{sqrt(32)}}}
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x = {{{(-b +- sqrt(b^2-4ac))/(2a)}}} becomes x = {{{(2 +- sqrt(32))/2}}} which becomes x = 3.828427125 or x = -1.828427125.
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The graph of our equation is {{{x^2 - 2x - 3}}} with the restriction that {{{-1.828427125 <= x <= 3.828427125}}}.
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a graph of our equation would look like this:
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{{{graph(600,600,-3,5,-6,6,x^2-2x-3,4,1000/(x+2),1000/(x-4))}}}
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you can see that the lower limit for x will be around -1.828... and the upper limit for x will be around +3.828... and that anything in between is good.
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those horizontal and vertical lines are just there to let you see the x values and the y values easier.  they are not part of the quadratic equation.