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_(-I,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_(-I,128]