Question 148405
If we plot the points we get


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


Now draw a curve through these points to get the equation:


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


From the graph, we can see that the degree is even (since both ends of the graph go in the same y direction). Also, we can see that the leading coefficient is positive (since the graph opens upward). Finally, we can see that there are two zeros (since there are two x-intercepts).


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Since we have all of this info, we don't need the equation. However, if you do need the equation, this is how you would find it:


From the graph, we can see that the vertex is (-1,-3). 



{{{y=a(x-h)^2+k}}} Start with the general vertex equation.



{{{y=a(x--1)^2-3}}} Plug in {{{h=-1}}} and {{{k=-3}}}



{{{y=a(x+1)^2-3}}} Simplify.


Now let's plug in another point. Let's plug in (1,-1)


{{{-1=a(1+1)^2-3}}} Plug in {{{x=1}}} and {{{y=-1}}}



{{{-1=a(2)^2-3}}} Add



{{{-1=4a-3}}} Square 2 to get 4.



{{{0=4a-3+1}}} Add {{{1}}} to both sides.



{{{-4a=-3+1}}} Subtract {{{4a}}} from both sides.



{{{-4a=-2}}} Combine like terms on the right side.



{{{a=(-2)/(-4)}}} Divide both sides by {{{-4}}} to isolate {{{a}}}.



{{{a=1/2}}} Reduce.


So this means that the equation is {{{y=(1/2)(x+1)^2-3}}}


{{{y=(1/2)(x^2+2x+1)-3}}} FOIL



{{{y=(1/2)x^2+(1/2)2x+(1/2)1-3}}} Distribute



{{{y=(1/2)x^2+x+1/2-3}}} Multiply



{{{y=(1/2)x^2+x-5/2}}} Combine like terms.



So the equation that goes through the points is {{{y=(1/2)x^2+x-5/2}}} 


From the equation, we can see that the degree of the polynomial is even. Also, we can see that the leading coefficient is positive since {{{a=1/2}}}. Finally, if we use the quadratic formula, we'll find that the polynomial {{{y=(1/2)x^2+x-5/2}}} has 2 zeros.