Question 148407
# 26


First let's plot the points


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/points3.jpg" alt="Photobucket - Video and Image Hosting">



Now let's draw a quadratic through these points (let me know if any other type of polynomial needs to be drawn otherwise)


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/graphthroughpoints3.jpg" alt="Photobucket - Video and Image Hosting">




If you need the equation of the quadratic, simply set up the system of equations



{{{system(4a-2b+c=4,5.76a+2.4b+c=-2.6176,c=-4)}}}


note: let me know if you need help with the setup


Now use either your graphing calculator or other software to solve for "a","b", and "c" simultaneously through the use of matrices. The solution that I get is 


a = 1.04, b = -1.92 and c = -4


So the equation that goes through these points is {{{y=1.04x^2-1.92x-4}}}



<b>Is the degree of the polynomial odd or even?</b>

Notice how as x approaches negative infinity (ie x goes to the left), then y approaches infinity. Also as x approaches positive infinity (ie x goes to the right), then y approaches infinity. So this tells us that the degree of the polynomial is even. Also, from the equation {{{y=1.04x^2-1.92x-4}}}, we can see that the degree is 2, which is even.



<b>Is the leading coefficient of the polynomial positive or negative?</b>


Since the parabola is opening upward, this means that the leading coefficient is positive. Also, since the leading coefficient is 1.04, this also tells us that the leading coefficient is positive.




<b>Number of zeros? </b>

From the graph, we can see that there are two zeros. Also, if you use the quadratic equation on {{{y=1.04x^2-1.92x-4}}}, you'll find that the polynomial has 2 zeros.