![]() Īlternative solution for the second part of the solution:Īfter figuring out that the expression under the radical is always positive (part a), we can solve the radical and therefore denominator for the least possible value (minimum value). That means the set of all real numbers is the domain of the f(x) and the correct answer is. The radical is always positive and denominator is never equal to zero, so the f(x) is defined for all real values of x. There is no real value for x that will fit this equation. cannot be a negative number.ī) Set the denominator of the fractional function equal to zero and look for probable x values: There is no real value for x that will fit this equation, because any real value square is a positive number i.e. In order to find the impossible values of x, we should:Ī) Set the equation under the radical equal to zero and look for probable x values that make the expression inside the radical negative: The domain is defined as the set of possible values for the x variable. In this case, is 3 and is 2, so we get slopes of and. Remember that these slopes always come in pairs, with one being positive and the other being negative. For a hyperbola, the slopes of the asymptotes can be found by dividing by (remember to always put the vertical value,, above the horizontal value, ). The center is always found at, which in this case is. Therefore, this is a horizontal hyperbola. If the term is positive, the hyperbola opens vertically. In other words, if the term is positive, the hyperbola opens horizontally. ![]() In a hyperbola, the squared term with a positive coefficient represents the direction in which the hyperbola opens. one is negative and the other is positive), this equation must be a hyperbola, not an ellipse. Because the coefficients in front of the squared variables are different signs (i.e. Because there are two squared variables ( and ), this equation cannot be a parabola. The first step is to determine the type of conic section this equation represents. ![]() Luckily, this equation is already in standard form: First, we need to make sure the conic section equation is in a form we recognize.
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