Question 784564
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Terminology problem: The equation specified in step h below is NOT a quadratic <i><b>equation</b></i>.  It is a quadratic <i><b>function</b></i>.  *[tex \Large -x^2\ +\ 2x\ +\ 8\ =\ 0] would be a quadratic <i><b>equation</b></i>.  Quadratic equations have solutions over the complex numbers, but they do not have graphs, vertices, axes of symmetry, intercepts, domains, or ranges.  Quadratic functions have all these things.


Given the function:


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


For your particular given function: 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -1\ \ ], *[tex \LARGE b\ =\ 2\ \ ], and *[tex \LARGE c\ =\ 8] 


The *[tex \Large x]-coordinate of the vertex is given by *[tex \Large x_v\ =\ \frac{-b}{2a}] and the *[tex \Large y]-coordinate of the vertex is given by *[tex \Large y_v\ =\ f(x_v)\ =\ a(x_v)^2\ +\ b(x_v)\ +\ c]. So the vertex is *[tex \Large \left(x_v,\,y_v\right)].


The axis of symmetry is *[tex \Large x\ =\ x_v]


The *[tex \Large y]-coordinate of the *[tex \Large y]-intercept is found by substituting zero for *[tex \Large x], hence the *[tex \Large y]-coordinate of the *[tex \Large y]-intercept is *[tex \Large c], and therefore the *[tex \Large y]-intercept is the point *[tex \Large \left(0,\,c\right)].


The *[tex \Large x]-coordinates of the *[tex \Large x]-intercepts, if they exist, are the zeros of the function.  First, determine existence.  If *[tex \Large b^2\ -\ 4ac\ \geq\ 0], then the intercepts exist.  Otherwise not.  Hint:  If *[tex \Large a] and *[tex \Large c] have opposite signs, then there are always two real and unequal zeros.


If the intercepts exist, then the *[tex \Large x]-coordinates of the *[tex \Large x]-intercepts are found by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ \frac{-b\ -\ \sqrt{b^2\ -\ 4ac}}{2a}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_2\ =\ \frac{-b\ +\ \sqrt{b^2\ -\ 4ac}}{2a}] 


And the *[tex \Large x]-intercepts are the points *[tex \Large \left(x_1,\,0\right)] and *[tex \Large \left(x_2,\,0\right)].  In the case of a perfect square trinomial that has one zero with a multiplicity of 2, there is only one intercept point, and that point is identical to the vertex, see part a.


The domain of all polynomial functions is the set of real numbers, that is: *[tex \Large \text{dom}f\ =\ \left\{x\ \in\ \mathbb{R}\right\}]


If the lead coefficient is positive, then the graph opens upward and the *[tex \Large y]-coordinate of the vertex, *[tex \Large y_v], is the minimum value of the function.  In this case, the function would have no maximum value since *[tex \Large f(x)] increases without bound as *[tex \Large x] either increases or decreases without bound.  The range in this case would be *[tex \Large \text{ran}f\ =\ \{y\ \in\ \mathbb{R}\ |\, y\ \geq \ y_v\}]


If the lead coefficient is negative, then the graph opens downward and the *[tex \Large y]-coordinate of the vertex, *[tex \Large y_v], is the maximum value of the function.  In this case, the function would have no minimum value since *[tex \Large f(x)] decreases without bound as *[tex \Large x] either increases or decreases without bound. The range in this case would be *[tex \Large \text{ran}f\ =\ \{y\ \in\ \mathbb{R}\ |\, y\ \leq \ y_v\}]


You can do your own arithmetic. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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