algebra systems of linear equations

[deniser7]

Find the equation of the parabola that passes through the given points....?
(4,-54),(-2,-6),(-3,-19)
write in form - y=ax[2]+bx+c

 

[mmm4444bot]

We form a system of three equations by substituting the given values for x and y into the Standard Form of a Quadratic Equation, one pair at a time:

y = Ax^2 + Bx + C

You need to solve the resulting system of three equations for the parameters A, B, and C.

You shared no thoughts or work thus far, so I have no idea what you expect to get here.

If you need more assistance, please provide specific details about where you're stuck or what you don't understand.

Cheers ~ Mark

 

[deniser7]

I understand that the new problems formed will be:
-54=16a+4b+c
-6=4a-2b+c
-19=9a-19b+c
but i can not get them to cancel out
please show me the steps

 

[mmm4444bot]

-54=16a+4b+c

-6=4a-2b+c

-19=9a-19b+c

Thank you, for sharing your thoughts. Fix the equation in red above, and you're good to go.


but i can not get them to cancel out

"them" meaning the coefficients ?

I mean, to me, this statement implies that you're talking about the Elimination Method, but I'm not certain what you're trying.

There are many different approaches, for finding the values of the coefficients A, B, and C.

Here is the beginning of one way.

Since each equation contains the same constant term C, we can obtain a system of two equations with A and B by solving each equation for C, followed by equating pairs of expressions for C.

I mean, if we make three new equations, like this:

C = first expression containing A and B

C = second expression containing A and B

C = third expression containing A and B

then all three expressions represent the same value, and we can form new equations by pairing, like this:

first expression = second expression

second expression = third expression

These last two equations will form a system of two equations with A and B.

I'll do the first pair, in your exercise, for you. Solving each of the first two equations for C, I get:

C = -16A - 4B - 54

C = -4A + 2B - 6

Both expressions equal C, so they are obviously the same.

-16A - 4B - 54 = -4A + 2B - 6

This simplifies to 12A + 6B = -48.

Please show me what you get, using the second pair of equations.

Then, we can use the Elimination Method, to solve for A and B, if that's what you want to do.

Otherwise, just show me what method you're trying; I need more details. Is your class covering a particular method?