Question 321729
<font face="Garamond" size="+2">


The *[tex \Large x] coordinates of the *[tex \Large x]-intercepts of a polynomial function are the zeros of the function.  The Fundamental Theorem of Algebra says that a degree *[tex \Large n] polynomial equation has *[tex \Large n] roots, hence the described parabola, being a 2nd degree polynomial, has exactly 2 roots. If *[tex \Large \alpha] is a zero of a polynomial, then *[tex \Large x\ -\ \alpha] is a factor of the polynomial.


The zeros are -1 and 2, hence the factors are *[tex \Large x\ +\ 1] and *[tex \Large x\ -\ 2].  Therefore, the general form of the polynomial is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ k(x\ +\ 1)(x\ -\ 2)\ =\ k\left(x^2\ -\ x\ -\ 2\right)]


Given the initial condition (based on the coordinates of the *[tex \Large y]-intercept, *[tex \Large (0,-16)]), allows us to calculate *[tex \Large k]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ kx^2\ -\ kx\ -\ k2]


Since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(0)\ =\ -16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k(0)^2\ -\ k(0)\ -\ k2\ =\ -16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ 8]


And the polynomial is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ 8x^2\ -\ 8x\ -\ 16]


An equation of the line that passes through *[tex \LARGE \left(-5,-10\right)] and *[tex \LARGE \left(4,30\right)] is given by the two-point form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ \left(\frac{y_1\ -\ y_2}{x_1\ -\ x_2}\right)(x\ -\ x_1) ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ 10\ =\ \left(\frac{-10\ -\ 30}{-5\ -\ 4}\right)(x\ +\ 5) ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{40}{9}x\ -\ \frac{290}{9} ]


Set the two RHSs equal:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x^2\ -\ 8x\ -\ 16\ =\ -\frac{40}{9}x\ -\ \frac{290}{9}]


Real roots of the above equation, if any exist, will be the *[tex \Large x] coordinates of the ordered pair(s) that represent the solution set of the system of equations.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 72x^2\ -\ 72x\ -\ 144\ =\ -40x\ -\ 290]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 72x^2\ -\ 32x\ +\ 146\ =\ 0]


But *[tex \LARGE 32^2\ -\ 4(72)(146)\ =\ -41024\ <\ 0]


So the roots are a conjugate pair of complex numbers with a non-zero imaginary part.  Therefore there are no real roots, and therefore no points of intersection of the two graphs.


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