Question 245946
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Hint for next time:  show square roots like this:  sqrt(x) whatever is inside the square root sign is inside the parentheses.  show plus or minus like this:  +\- 


So you got yourself to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2\ = \frac{2}{3}]


and then you took the square root of both sides to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \pm\sqrt{\frac{2}{3}}]


So far, so good.  But, you need to simplify further.  A radical in a denominator is not simplest form.  First apply the rule that the square root of a quotient is the quotient of the square roots.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{\frac{a}{b}}\ =\ \frac{\sqrt{a}}{\sqrt{b}}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \pm\frac{\sqrt{2}}{\sqrt{3}}]


Now multiply your fraction by 1 in the form of *[tex \Large \frac{\sqrt{3}}{\sqrt{3}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \pm\frac{\sqrt{2}}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{3}}\right)\ =\ \pm\frac{\sqrt{6}}{3}]


And now you are in simplest form, though you might not think so just to look at it.  But the rule is:  No matter how ugly the numerator gets, a fraction is simpler if it has no radical in the denominator.


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You are right.  There are no integer factors for the given quadratic.  So you need to either Complete the Square (which will get messy because the first degree coefficient is odd), or use the quadratic formula:


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


where *[tex \Large a], *[tex \Large b], and *[tex \Large c] are the high order, first degree, and constant coefficients of the quadratic in standard form.  For your problem:  *[tex \Large a\ =\ 1], *[tex \Large b\ =\ -3], and *[tex \Large c\ =\ 7]


Since the discriminant, *[tex \Large b^2\ -\ 4ac\ <\ 0], the solution will be a conjugate pair of complex roots of the form *[tex \Large \alpha\ \pm\ \beta i] where *[tex \Large i] is the imaginary number defined as *[tex \Large i^2\ = -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(-3)\ \pm\ \sqrt{(-3)^2\ -\ 4(1)(7)}}{2(1)}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{3\ \pm\ i\sqrt{19}}{2}]



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For any function of the form


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


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


The *[tex \LARGE y]-coordinate of the vertex is the value of the function at the  *[tex \LARGE x]-coordinate of the vertex, namely: *[tex \LARGE f\left(\frac{-b}{2a}\right)]


This is a parabola, so there is only one line of symmetry, namely the vertical line that passes through the vertex, so the equation is:


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


If the lead coefficient is positive, then the parabola opens upward.  If the parabola opens upward then quite obviously the *[tex \LARGE y]-coordinate of the vertex is the minimum of the function.  In this case, the function has no maximum.


If the lead coefficient is negative, then the opposite is true:  The parabola opens downward, the value of the function at the vertex is a maximum, and there is no minimum.


So, for your particular problem, you need to first calculate the value of *[tex \LARGE \frac{-(5)}{2(2)}], and then calculate the value of the function at that value:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f\left(\frac{-(5)}{2(2)}\right)\ =\ 2\left(\frac{-(5)}{2(2)}\right)^2\ +\ 5\left(\frac{-(5)}{2(2)}\right)\ -\ 1]


That will give you the vertex and the axis of symmetry.


Evaluate


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f\left(0\right)\ =\ 2\left(0\right)^2\ +\ 5\left(0\right)\ -\ 1]


To determine the point *[tex \Large \left(0,\,f(0)\right)] where the graph intersects the *[tex \Large y]-axis.


Solve the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 5x\ -\ 1\ =\ 0]


to find the points where the graph intersects the *[tex \Large x]-axis.


Use symmetry to find other points.  If *[tex \Large x_v] is the *[tex \Large x]-coordinate of the vertex, *[tex \Large f\left(x_v\ +\ a\right)\ =\ f\left(x_v\ -\ a\right)]


Plot all of your points and draw a smooth curve.



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