Question 189058
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The vertex of a parabola is a point on the plane, so you have to determine the <i>x</i> and <i>y</i> coordinates in order to define that point.  Fortunately, we have a formula available for just this purpose.


The <i>x</i>-coordinate of the vertex of a parabola whose equation is in the form:


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


is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_V = \frac{-b}{2a}]


For your given function,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x) = x^2 + 10x + 14]


<i>a</i> = 1 and <i>b</i> = 10, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_V = \frac{-10}{2(1)} = -5]


Now that we know the <i>x</i>-coordinate of the vertex, we can find the value of the function at that <i>x</i> value to compute the <i>y</i>-coordinate of the vertex, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_V\ =\ f(x_V)\ =\ f(-5)\ =\ (-5)^2 + 10(-5) + 14\ =\ 25 - 50 + 14\ =\ -11]


So, the vertex is the point (-5,-11)


For your other problem, perform the same process to discover the coordinates of the vertex.  You will find that the value of the function at the vertex (same thing as the <i>y</i>-coordinate of the vertex), given properly performed arithmetic, is something other than 20.  As to whether it is a maximum or minimum, you can tell by the sign on the lead coefficient.


If the lead coefficient is > 0, then the parabola opens upward and the value of the function at the vertex is a minimum.  If the lead coefficient is < 0, then it is the opposite case, i.e the graph opens downward and the vertex is a maximum.  And, of course, if the lead coefficient = 0, then you don't have a parabola at all, rather you have a straight line.


For your second problem, the lead coefficient, the <i>a</i> in *[tex \large f(x) = ax^2 + bx + c], is -1, so you have a graph that opens downward and the vertex is a maximum.


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