Question

What Is The Equation, In Standard Form, Of A Parabola That Models The Values In The Table?
X -2, 0, 4
Y 1, 5, -59

Answer 1

Standard form:\[y = ax^2+bx+c\]Plug each of the values you are given into the equation. This will give you three equations in three unknowns \((a,b,c)\), which you solve to get the coefficients for your equation.

For example, the first point is \((-2,1)\) so
\[1=a(-2)^2+b(-2)+c\]\[1 = 4a-2b+c\]

Answer 2

Thank You c:

Answer 3

What did you get for your final answer?

Answer 4

Wait Is That All I Do? And Im Still Working It Out.

Answer 5

Find the 3 values \(a,b,c\) by solving the system of 3 equations, then plug them into the template:
\[y = ax^2 + bx + c\]

No rush, just offering to check your answer when you've got it done.

Answer 6

Okkie Dokkie Thank You and Ill Let You Know.

Answer 7

Okay I Got:
1=4a-2b+c
5=a+b+c
-59=16a+4b+c
Is That Right?

Answer 8

No, the second one isn't correct. Mind showing me your work?

Answer 9

I think you probably put in \(x=1\) instead of \(x =0\)...

Answer 10

I Did This:
5=a(0)^2+b(0)+c

Answer 11

Okay, how do you get from that to \[5 = a+b+c\]?

Answer 12

\[a(0)^2 =\]\[b(0)=\]?

Answer 13

Hmm Now I See How That Doesnt Make Sense. Wouldnt That Be a^2?

Answer 14

Sorry, I didn't see the notification that you'd responded :-(

No, it isn't \(a^2\), the formula is \(y = ax^2 + bx + c\)

We wouldn't want to have to square, cube, etc. the coefficients to match the terms, because if we needed a negative number for a term with an even exponent, we would need to use a complex number as our coefficient.

For example, if we had a term \(-x^2 = -1x^2\), if we used \(a^2\) to match \(x^2\), then we would have to find a value of \(a\) such that \(a^2= -1\). There aren't any real numbers that work, only complex ones.

Answer 15

In any case, \[5 = a(0)^2+b(0) + c\]simplifies to \[5 = 0a+0b+c\]\[5 = c\]

(and for that matter, so does your proposed variation!)
\[5 = a^2(0)^2 + b(0) +c \]\[5=0a^2+0b+c\]\[5 = c\]