Question 618833
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I'm not sure what sort of new math folderol "build a pattern using tiles" is supposed to be, but yes, plug each of the given *[tex \LARGE x] values into your given function and report the function value for each.


The *[tex \LARGE y]-intercept is the point with a zero *[tex \LARGE x]-coordinate and the value of the function when *[tex \LARGE x\ =\ 0] for a *[tex \LARGE y]-coordinate.


Set the function equal to zero and then solve the resulting quadratic equation for the two roots.  The *[tex \LARGE x]-intercepts will be the two points where the *[tex \LARGE x]-coordinate is equal to one or the other of the roots and the *[tex \LARGE y]-coordinate is zero.


Find the *[tex \LARGE x]-coordinate of the vertex using the formula *[tex \LARGE x_v\ =\ \frac{-b}{2a}] where *[tex \LARGE a] is the lead coefficient and *[tex \LARGE b] is the coefficient on the first degree term, as in *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c]


The *[tex \LARGE y]-coordinate of the vertex is the value of the function at the *[tex \LARGE x]-coordinate of the vertex.  Hint: You have already calculated it in part B of the problem.


Since the lead coefficient is positive, the parabola opens upward.  Hence the vertex is a minimum of the function.  Therefore the lower limit of the range is the value of the function at the vertex, i.e., the number you calculated in the previous step.  Since this is a parabola and the domain has not been otherwise restricted, there is no upper bound to the range.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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