Question 190301
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x) = x^2 + x - 2 ] 


Since there are no real number values for which this function is undefined the domain is all real numbers or *[tex \LARGE (-\infty,\infty)].


Since this is a quadratic polynomial, the graph is a parabola.  Since the lead coefficient is positive, the parabola opens upward.  Let's table discussion of the Range until we have identified the vertex.


The <i>x</i>-coordinate of the vertex of a parabola described by a function in the form:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \phi(x) = ax^2 + bx + c ]


is given by: *[tex \LARGE x_V = \frac{-b}{2a}]


so for your function: *[tex \LARGE x_V = \frac{-1}{2(1)} = -\frac{1}{2}]


The <i>y</i>-coordinate of the vertex of such a parabola is given by


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_V\ =\ f(x_V)\ =\ \left(-\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) - 2 = \frac{1}{4} - \frac{1}{2} - 2 = -\frac{9}{4}]


So the vertex is the point *[tex \LARGE \left(-\frac{1}{2},-\frac{9}{4}\right)]


Since this is a parabola opening upward, the <i>y</i>-coordinate of the vertex represents the minimum value of the function.  And there is no maximum value, so we can now define the range as [*[tex \LARGE -\frac{9}{4}],*[tex \LARGE \infty]).  This also answers the question about whether there is a minimum or maximum -- and what the value of the minimum is.


The axis of symmetry is the vertical line defined by the equation *[tex \LARGE x = a] where <i>a</i> is the value of the <i>x</i>-coordinate of the vertex, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = -\frac{1}{2}]


The <i>y</i>-intercept is the value of the function when <i>x</i> equals zero, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(0) = (0)^2 + (0) - 2 =-2] 


The <i>x</i>-intercepts are the values of <i>x</i> that make the value of the function equal zero, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + x - 2 = 0] 


The trinomial factors, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x + 2)(x - 1) = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1 = -2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_2 = 1]


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John
*[tex \LARGE e^{i\pi} + 1 = 0]
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