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Mathtrouble12
What is the domain and range of the quadratic equation y = –2x^2– 32x – 126?
y=-2x^(2)-32x-126 Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. -2x^(2)-32x-126=y To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side. -2x^(2)-32x-y-126=0 Multiply each term in the equation by -1. -2x^(2)*-1-32x*-1-y*-1-126*-1=0*-1 Simplify the left-hand 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(y-128)))/(2(2)) Simplify the denominator of the quadratic formula. x=(-32\2~(-2(y-128)))/(4) First, solve the + portion of \. x=(-32+2~(-2(y-128)))/(4) Simplify the expression to solve for the + portion of the \. x=(-16+~(-2(y-128)))/(2) Next, solve the - portion of \. x=(-32-2~(-2(y-128)))/(4) Simplify the expression to solve for the - portion of the \. x=(-16-~(-2(y-128)))/(2) The final answer is the combination of both solutions. x=(-16+~(-2(y-128)))/(2),(-16-~(-2(y-128)))/(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(y-128))<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)-32x-126 is equal to the range of f(y)=((-16+~(-2(y-128))))/(2). Range: y<=128_(-<Z>I<z>,128]
domain: 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 \[(-\infty,2]\]