Find the equation for a parabola given 3 points

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To find the equation of a parabola passing through the points (1,-2), (-3,10), and (4,31), a system of linear equations is established based on the general form of a parabola. The equations are set up as follows: -2 = a + b + c, 10 = 9a - 3b + c, and 31 = 16a + 4b + c. The discussion involves using triple elimination, which is a method for solving these equations, to isolate the coefficients a, b, and c. The correct solution for the parabola is found to be y = 2x^2 + x - 5. The conversation highlights the challenges faced by high school students in understanding quadratic equations and solving systems of equations.
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Homework Statement



Find an equation for a parabola that passes through the following points. To solve you must use triple elimination, you may check you answer any way you wish

Homework Equations





The Attempt at a Solution


okay so the points given were (1,-2) (-3,10) and (4,31)
I got to the point of making equations to olve with but then my brain hit a wall. What do i need to do next?
-2=a+b+c
10=9a -3b +c
and
31= 16a +4b +c
 
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i really am not good at quadratics
 
Doodledawg said:

The Attempt at a Solution


okay so the points given were (1,-2) (-3,10) and (4,31)
I got to the point of making equations to olve with but then my brain hit a wall. What do i need to do next?
-2=a+b+c
10=9a -3b +c
and
31= 16a +4b +c

This is a system of linear equations. There are several ways to solve for a,b, and c. One way is to use linear algebra and set up a matrix equation "Ax=b". Where A is the matrix with coefficients, x is vector [a b c] and b is vector [-2 10 31]
\[ \left( \begin{array}{ccc}<br /> 1 &amp; 1 &amp; 1 \\<br /> 9 &amp; -3 &amp; 1 \\<br /> 16 &amp; 4 &amp; 1 \end{array} \right)\]
Now, solve for x
 
What you found is now a system of three simultaneous equations in three unknowns (which you've set up correctly), those unknowns being the coefficients of the equation for your parabola. Are you comfortable with solving such systems?
 
im not very comforable with them
 
okay i got down to b=2a-3
but i don't know where to go from here
 
konthelion said:
This is a system of linear equations. There are several ways to solve for a,b, and c. One way is to use linear algebra and set up a matrix equation "Ax=b". Where A is the matrix with coefficients, x is vector [a b c] and b is vector [-2 10 31]
\[ \left( \begin{array}{ccc}<br /> 1 &amp; 1 &amp; 1 \\<br /> 9 &amp; -3 &amp; 1 \\<br /> 16 &amp; 4 &amp; 1 \end{array} \right)\]
Now, solve for x
I have to use triple elimination
 
okay i got the right answer with the matrices:
y=2x^2+x-5
but how do i find this through triple elimination?
 
Doodledawg said:
okay i got the right answer with the matrices:
y=2x^2+x-5
but how do i find this through triple elimination?
Could you explain what a triple elimination is? I've never heard this term before. Is this another name for Gaussian elimination?
 
  • #10
no it's the classic way of doing things, we do it in high school.

Ok, so you have 3 equations:

[1]. -2=a + b +c
[2]. 10=9a -3b +c
[3]. 31= 16a +4b +c

Eliminate a by using 9 * [1] - [2] and then 16 * [1] - [3]

(-18 = 9a + 9b + 9c) - (10 = 9a - 3b + c) ---> -28 = 12b + 8c [4]
(-32 = 16a + 16b + 16c) - (31= 16a +4b +c) ---> -63 = 12b + 15c [5]

eq [4] - eq [5], should be left with c now.

Now solve for c

Plug c back into [4]

Solve for b

Now use eq [1], plug back b and c

Solve for a
 
  • #11
yeah ok thx
im actually in high school and feel like a total idiot but thanks for helping me
i much appreciate it
 
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