Question 311070
<font face="Garamond" size="+2">


You need to first write the equation of the parabola.  We have a vertex at *[tex \Large (-1,2)] and a *[tex \Large y]-intercept at *[tex \Large (0,-3)].  The vertex is at a distance of 1 from the *[tex \Large y]-axis.  Therefore, because of symmetry, you will find a point on the parabola with a *[tex \Large y]-coordinate of -3 at a distance of 1 on the other side of the vertex, namely the point *[tex \Large (-2,-3)].


Now that we have three points, we can go about finding the equation of the parabola.


The general equation of a parabola is:


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


Substituting the values from our three points we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(0)^2\ +\ b(0)\ +\ c\ =\ -3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-1)^2\ +\ b(-1)\ +\ c\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-2)^2\ +\ b(-2)\ +\ c\ =\ -3]


A little arithmetic results in the following system of three linear equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ -3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ +\ c\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ +\ c\ =\ -3]


Solve the system.  First substitute -3 for *[tex \Large c]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ =\ 0]


Multiply the the second equation just above by *[tex \Large -\frac{1}{2}]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2a\ +\ b\ =\ 0]


Add the resulting two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -a\ =\ 5\ \ \Rightarrow\ \ a\ =\ -5]


And finally by substitution:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ -10]


Hence, the desired function is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -5x^2\ -\ 10x\ -\ 3]


Find the roots with the quadratic formula since there are no rational factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(-10)\ \pm\ \sqrt{(-10)^2\ -\ 4(-5)(-3)}}{2(-5)} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{10\ \pm\ \sqrt{40}}{-10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{10\ \pm\ 2\sqrt{10}}{-10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{5\ \pm\ \sqrt{10}}{-5}]


Therefore the *[tex \Large x]-intercepts are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{5\,-\,\sqrt{10}}{-5},\,0\right)\ \approx\ \left(-1.632,0\right)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{5\,+\,\sqrt{10}}{-5},\,0\right)\ \approx\ \left(-0.368,0\right)]



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