Question 698014
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For any quadratic of the form *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c], the *[tex \LARGE x]-coordinate of the vertex is found by:


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


and the *[tex \LARGE y]-coordinate of the vertex is found by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v = \rho(x_v) = \rho\left(\frac{-b}{2a}\right)\ =\ a\left(\frac{-b}{2a}\right)^2\ +\ b\left(\frac{-b}{2a}\right)\ +\ c]


An equation of the axis of symmetry is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ x_v]


which is to say the vertical line that passes through the vertex.


To find the *[tex \LARGE y]-intercept, set *[tex \LARGE x\ =\ 0], then calculate the resulting value of the function, *[tex \LARGE y\ =\ \beta].  Then the *[tex \LARGE y]-intercept is the point *[tex \LARGE (0,\beta)]


The *[tex \LARGE x]-intercepts are the points *[tex \LARGE \left(\alpha_1,0\right)] and *[tex \LARGE \left(\alpha_2,0\right)] where *[tex \LARGE \alpha_1] and *[tex \LARGE \alpha_2] are the real number zeros of the function.  The first thing to do is to determine whether the function has any real zeros.


<u>Nature of zeroes</u>


Find the Discriminant, *[tex \LARGE \Delta\ =\ b^2\ -\ 4ac] and evaluate the nature of the zeroes as follows:


No calculation quick look:  If the signs on *[tex \Large a] and *[tex \Large c] are opposite, then *[tex \LARGE \Delta > 0] guaranteed.


*[tex \LARGE \Delta > 0 \ \ \Rightarrow\ \] Two real and unequal zeroes. If *[tex \LARGE \Delta] is a perfect square, the quadratic factors over *[tex \LARGE \mathbb{Q}] (the rationals).


*[tex \LARGE \Delta = 0 \ \ \Rightarrow\ \] One real zero with a multiplicity of two.  That is to say that the trinomial is a perfect square and has two identical factors.  Presuming rational coefficients, the zero will be rational as well.


*[tex \LARGE \Delta < 0 \ \ \Rightarrow\ \] A conjugate pair of complex number zeroes of the form *[tex \LARGE a \pm bi] where *[tex \LARGE i] is the imaginary number defined by *[tex \LARGE i^2 = -1].  In this case, a quadratic function does NOT intercept the *[tex \LARGE x]-axis.


To find the domain, determine the set of real numbers for which the function is defined.  Hint:  The domain of ALL polynomial functions is the set of real numbers.  In interval notation, *[tex \LARGE \left(-\infty,\infty\right)].  In set notation:  *[tex \LARGE \text{dom}\{\rho(x)\}\ =\ \{x\,\in\,\mathbb{R}\}]


To find the range, determine whether the *[tex \LARGE y]-coordinate  is of the vertex is a minimum or a maximum value of the function.  If *[tex \LARGE a\ >\ 0], then the graph of the quadratic is a parabola that opens upward, meaning that the vertex is at the bottom of the graph, so the *[tex \LARGE y]-coordinate of the vertex is a ______________.  (You fill in the blank).  If *[tex \LARGE a\ <\ 0], then we have the opposite case.  If the vertex represents a minimum, then the lower limit of the range is the *[tex \LARGE y]-coordinate of the vertex and the upper limit increases without bound. If the vertex represents a maximum, then the *[tex \LARGE y]-coordinate represents the upper limit of the range and the lower limit decreases without bound.  The other Hence, the range is represented either as an interval or a set.

Interval Notation:  (*[tex -\infty,y_v]] or [*[tex y_v,\infty])


Sets:


*[tex \LARGE \text{rng}\{\rho(x)\}\ =\ \{y,\in\,\mathbb{R}\,=\,\rho(x)\ |\ y_v\ \leq\ y\ <\ \infty\}]


or


*[tex \LARGE \text{rng}\{\rho(x)\}\ =\ \{y,\in\,\mathbb{R}\,=\,\rho(x)\ |\ -\infty\ <\ y\ \leq\ y_v\}]


Regions of increase or decrease.  Start with a value for *[tex \LARGE x] that is strictly less than *[tex \LARGE x_v].  Determine the value of the function at this selected *[tex \LARGE x] value.  Compare the calculated function value with *[tex \LARGE y_v].  If the calculated function value is LARGER than *[tex \LARGE y_v], then the function is DECREASING on the interval (*[tex -\infty,x_v]].  In this case, the function is, perforce, increasing on the other half-plane interval, to wit: [*[tex x_v,\infty]).  On the other hand, if the calculated function value when the selected value of *[tex \LARGE x] is SMALLER than *[tex \LARGE y_v], then the function is INCREASING on the side of the graph to the left of the axis of symmetry and decreasing on the other side.


<u>Graphing</u>


1.  Plot the vertex.


2.  Plot the *[tex \LARGE y]-intercept


3.  Plot the *[tex \LARGE x]-intercepts if they exist


4.  Use symmetry to find a point that corresponds to the *[tex \LARGE y]-intercept.  Determine the distance along the *[tex \LARGE x]-axis between the *[tex \LARGE y]-axis and the axis of symmetry.  Move that same distance on the OTHER side of the axis of symmetry.  At that *[tex \LARGE x]-value, there will be a point that has the same function value (that is the same *[tex \LARGE y]-value] as the *[tex \LARGE y]-value of the *[tex \LARGE y]-intercept.  In general, points on the graph of the parabola equidistant from the axis of symmetry have identical function (or *[tex \LARGE y]) values.


5.  Pick another *[tex \LARGE x] value that has not been used.  Calculate the function value and plot the point.  Use symmetry to find the corresponding point on the other side of the axis of symmetry.


6.  When you have enough points plotted to suit yourself, draw a smooth curve through the plotted points.


<u>The importance of reading instructions</u>.  One of the reasons you may be having difficulty with mathematics is your demonstrated inability to either read or follow (or both) written instructions.  Before you post in the future, please take some time to read and understand the instructions for posting on this site.


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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