Question 238497
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ x^2\ +\ 4x\ -\ 45]


is a parabola of the form 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c]


The lead coefficient is positive so the parabola opens upward.


The *[tex \Large x]-coordinate of the vertex is given by *[tex \Large -\frac{b}{2a}]


Then obviously the *[tex \Large y]-coordinate of the vertex must be *[tex \Large f\left(\frac{b}{2a}\right)]


The axis of symmetry is the vertical line *[tex \Large x\ =\ -\frac{b}{2a}]


The *[tex \Large x]-intercepts, if any exist, are the values of *[tex \Large x] that satisfy *[tex f(x)\ =\ 0]


The *[tex \Large y]-intercept is found by evaluating *[tex f(0)]


The domain of any polynomial function is all real numbers.  The range of a quadratic polynomial in the form


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c]


that opens upward is *[tex \Large \{y\,|\,y\,\in\,\R,\ y\ \geq\ f\left(\frac{b}{2a}\right)\}]


A parabola that opens upward is increasing to the right of the vertex, that is to say the interval <font size="+3">[*[tex \Large -\frac{b}{2a},\infty])<font size="+2">


Quite obviously the decreasing interval is <font size="+3">(*[tex \Large -\infty,-\frac{b}{2a}]]<font size="+2">


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