Question 1206777
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What is the range of y = -x^2 - 2x + 3
A. x ≤ 4
B. x ≤ -4
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<pre>
Function y(x) is a parabola opened downward.


The vertex is at  x = {{{x[max]}}} = = " {{{-b/(2a) }}}' = {{{- (-2)/(2*(-1))}}} = -1.


The maximum value of  y(x)  is at  x = -1:  {{{y[max]}}} = -(-1)^2 - 2*(-1) + 3 = -1 + 2 + 3 = 4.



In case (A),  the value  {{{x[max]}}} = -1  is in the domain;  THEREFORE,

the  <U>ANSWER</U>  to (A) is :  at x <=4,  the range of  y(x)  is  (-oo,{{{y[max]}}}] = (-oo,4].



In case (B), the value of {{{x[max]}}} = -1 is  out the domain of y(x);  THEREFORE,

    the  <U>ANSWER</U>  to (B) is :  at x <= -4,  the range of  y(x)  is  (-oo,y(-4)].

    y(-4) = -(-4)^2 - 2*(-4) + 3 = -16 + 8  + 3 = -5;  so,  the range of  y(x)  is  (-oo,-5].
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Solved.