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anonymous
 5 years ago
What is the domain and range of the quadratic equation y = –2x^2– 32x – 126?
anonymous
 5 years ago
What is the domain and range of the quadratic equation y = –2x^2– 32x – 126?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y=2x^(2)32x126 Since x is on the righthand side of the equation, switch the sides so it is on the lefthand side of the equation. 2x^(2)32x126=y To set the lefthand side of the equation equal to 0, move all the expressions to the lefthand side. 2x^(2)32xy126=0 Multiply each term in the equation by 1. 2x^(2)*132x*1y*1126*1=0*1 Simplify the lefthand side of the equation by multiplying out all the terms. 2x^(2)+32x+y+126=0*1 Multiply 0 by 1 to get 0. 2x^(2)+32x+y+126=0 Use the quadratic formula to find the solutions. In this case, the values are a=2, b=32, and c=1y. x=(b\~(b^(2)4ac))/(2a) where ax^(2)+bx+c=0 Substitute in the values of a=2, b=32, and c=1y. x=(32\~((32)^(2)4(2)(1y)))/(2(2)) Simplify the section inside the radical. x=(32\2~(2(y128)))/(2(2)) Simplify the denominator of the quadratic formula. x=(32\2~(2(y128)))/(4) First, solve the + portion of \. x=(32+2~(2(y128)))/(4) Simplify the expression to solve for the + portion of the \. x=(16+~(2(y128)))/(2) Next, solve the  portion of \. x=(322~(2(y128)))/(4) Simplify the expression to solve for the  portion of the \. x=(16~(2(y128)))/(2) The final answer is the combination of both solutions. x=(16+~(2(y128)))/(2),(16~(2(y128)))/(2) The domain of an expression is all real numbers except for the regions where the expression is undefined. This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0. All of these are undefined and therefore are not part of the domain. (2(y128))<0 Solve the equation to find where the original expression is undefined. y>128 The domain of the rational expression is all real numbers except where the expression is undefined. y<=128_(<Z>I<z>,128] The domain of the inverse of y=2x^(2)32x126 is equal to the range of f(y)=((16+~(2(y128))))/(2). Range: y<=128_(<Z>I<z>,128]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0domain: ALL REAL NUMBERS range: FROM SECOND COORDINATE OF VERTEX DOWN vertex is \[\frac{b}{2a}\] in this case you get 8 plug that in, get 2 range is \[(\infty,2]\]
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