Recall that each point is of the form (x,y). So for instance, the point (-1, 5) means x=-1 and y=5. This applies for each point given.



Also, remember that every quadratic can be represented as the equation:



{{{y=ax^2+bx+c}}} 



where 'a', 'b' and 'c' are real numbers. These values are usually known (and we solve for 'x'), but in this case, we must set up a system of equations to find these values. Note: it turns out that there is only one unique solution to this problem.


So....


For the point (-1,5) we know that x=-1 and y=5. Since this point lies on the quadratic, we know that if we plug in x=-1 into the unknown quadratic, we know that we'll get y=5. So the idea is to plug in the given values to find the unknown values.



Plug these values into the general equation {{{y=ax^2+bx+c}}} to get


{{{5=a(-1)^2+b(-1)+c}}}



Now square -1 to get 1, which means it will absorb into 'a', and simplify:



{{{5=a-b+c}}}



So the first equation we get is {{{5=a-b+c}}}



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Furthermore, since the parabola goes through (0,3) we can plug in x=0 and y=3 to get {{{3=a(0)^2+b(0)+c}}} which simplifies to {{{c=3}}}. Because we've already isolated 'c', we can use this and plug it into the first equation {{{5=a-b+c}}} to get {{{5=a-b+3}}}. Now solve for 'a' to get {{{a=b+2}}}. So whatever 'b' is, the value of 'a' will be 2 more than that.


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Finally, we see that the point (3,9) lies on the parabola. So just plug in x=3 and y=9 to get {{{9=a(3)^2+b(3)+c}}} and simplify: {{{9=9a+3b+c}}}



From here, we'll use the previously solved for variables 'a' and 'c' to find 'b':



{{{9=9a+3b+c}}} Start with the given equation.



{{{9=9(b+2)+3b+3}}} Plug in {{{a=b+2}}} and {{{c=3}}}



{{{9=9b+18+3b+3}}} Distribute



{{{9=12b+21}}} Combine like terms.



{{{9-21=12b}}} Subtract 21 from both sides.



{{{-12=12b}}} Combine like terms.



{{{-12/12=b}}} Divide both sides by 12 to isolate 'b'.



{{{-1=b}}} Reduce



So the value of 'b' is {{{b=-1}}}. Remember that we found that {{{a=b+2}}}. So {{{a=(-1)+2=1}}} which means {{{a=1}}}



So after everything is said and done, we find that {{{a=1}}}, {{{b=-1}}} and {{{c=3}}} giving us the quadratic {{{y=x^2-x+3}}}




Notice how the parabola {{{y=x^2-x+3}}} goes through the points (-1, 5), (0,3), and (3,9). So this visually confirms our answer. 



{{{ drawing(500, 500, -10,10,-5,15,
grid(1),
graph(500, 500, -10,10,-5,15, x^2-x+3),

circle(-1,5,0.08),circle(-1,5,0.10),
circle(0,3,0.08),circle(0,3,0.10),

circle(3,9,0.08),circle(3,9,0.10)


)}}} 


Graph of {{{y=x^2-x+3}}} through the points (-1, 5), (0,3), and (3,9)