SOLUTION: I need help with how to solve this:write a quadratic equation in standard form that passes through these points: (-1, 5), (0,3), (3,9). I know it's a systems of equations, but the

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> SOLUTION: I need help with how to solve this:write a quadratic equation in standard form that passes through these points: (-1, 5), (0,3), (3,9). I know it's a systems of equations, but the      Log On

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Question 235135: I need help with how to solve this:write a quadratic equation in standard form that passes through these points: (-1, 5), (0,3), (3,9).
I know it's a systems of equations, but the variables are confusing me... I got
5 = -A^2 - B + C
3 = C, and
9 = 3A^2 + 3B + C
I tried substituting 3 for c, but I don't know what to do from there
show work please and explain

Answer by jim_thompson5910(21667) About Me  (Show Source):
You can put this solution on YOUR website!
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%5E2%2Bbx%2Bc


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%5E2%2Bbx%2Bc to get

5=a%28-1%29%5E2%2Bb%28-1%29%2Bc


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


5=a-b%2Bc


So the first equation we get is 5=a-b%2Bc


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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%280%29%5E2%2Bb%280%29%2Bc 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%2Bc to get 5=a-b%2B3. Now solve for 'a' to get a=b%2B2. So whatever 'b' is, the value of 'a' will be 2 more than that.

--------------------------------------------


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%283%29%5E2%2Bb%283%29%2Bc and simplify: 9=9a%2B3b%2Bc


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


9=9a%2B3b%2Bc Start with the given equation.


9=9%28b%2B2%29%2B3b%2B3 Plug in a=b%2B2 and c=3


9=9b%2B18%2B3b%2B3 Distribute


9=12b%2B21 Combine like terms.


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


-12=12b Combine like terms.


-12%2F12=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%2B2. So a=%28-1%29%2B2=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%5E2-x%2B3



Notice how the parabola y=x%5E2-x%2B3 goes through the points (-1, 5), (0,3), and (3,9). So this visually confirms our answer.


+drawing%28500%2C+500%2C+-10%2C10%2C-5%2C15%2C%0D%0Agrid%281%29%2C%0D%0Agraph%28500%2C+500%2C+-10%2C10%2C-5%2C15%2C+x%5E2-x%2B3%29%2C%0D%0A%0D%0Acircle%28-1%2C5%2C0.08%29%2Ccircle%28-1%2C5%2C0.10%29%2C%0D%0Acircle%280%2C3%2C0.08%29%2Ccircle%280%2C3%2C0.10%29%2C%0D%0A%0D%0Acircle%283%2C9%2C0.08%29%2Ccircle%283%2C9%2C0.10%29%0D%0A%0D%0A%0D%0A%29

Graph of y=x%5E2-x%2B3 through the points (-1, 5), (0,3), and (3,9)