SOLUTION: What is the domain of f, range of f, vertex, axis of symmetry, x-intercept (s), y-intercept, determine whether it has a minimum or a maximum value and state that value for f(x)=x^2
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Question 190301: What is the domain of f, range of f, vertex, axis of symmetry, x-intercept (s), y-intercept, determine whether it has a minimum or a maximum value and state that value for f(x)=x^2+x-2
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Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
 = x^2 + x - 2 )
Since there are no real number values for which this function is undefined the domain is all real numbers or )
.
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 x-coordinate of the vertex of a parabola described by a function in the form:
 = ax^2 + bx + c )
is given by: 
so for your function: } = -\frac{1}{2})
The y-coordinate of the vertex of such a parabola is given by
\ =\ \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 )
Since this is a parabola opening upward, the y-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 [
,
). 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 
where a is the value of the x-coordinate of the vertex, so:
The y-intercept is the value of the function when x equals zero, so:
 = (0)^2 + (0) - 2 =-2)
The x-intercepts are the values of x that make the value of the function equal zero, so:
The trinomial factors, so:
(x - 1) = 0)
John
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