Can You Solve This Equation?

Question

I made a chart of values, and here's what it looks like:
X 1 2 3 4 5 6 7 8 9 10 11 12
Y 12 22 30 36 40 42 42 40 36 30 22 12

The graph is in the shape of a parabola. I would like the answer, but if you could also, please explain how you got it.

Thank You!

Answer 1

Let y = ax² + bx + c

When x = 1; y = a + b + c = 12 ............ (1)
When x = 2; y = 4a + 2b + c = 22 ........ (2)
When x = 3; y = 9a + 3b + c = 30 ........ (3)

On solving you get a = -1 b = 13 c = 0

ie y = 13x - x²

Agree with Antoine above

Answer 2

For a parabola, the general equation is
y = ax^2 + bx + c where a,b,c are constants
we can thus take 3 pairs of numbers to solve this since there are three unknowns, so:
12 = a + b + c eqtn 1
22 = 4a + 2b + c eqtn 2
30 = 9a + 3b + c eqtn 3

eqtn 2 - eqtn 1:
10 = 3a + b eqtn 4

eqtn 3 - eqtn 2:
8 = 5a + b eqtn 5

eqtn 5 - eqtn 4:
-2 = 2a so a = -1

working backwards, b = 13 and c = 0

so the parabola is y = -x^2 + 13x

Answer 3

The answer is y=-x^2+13x
If you have a graphing calculator you can use the quadratic regression function on it. Otherwise you need to solve a system of equations using three of the coordinate pairs. The reason you need three data pairs is the graph is a parabola which is a quadratic function. The general rule for polynomials is the degree plus 1 is the number of points needed to determine the equation.
The method of solving equations would be as follows:
12=1a+1b+1c 22=4a+2b+c 30=9a+3b+c
After subtraction the first equation from the second we get
10=3a+1b
We multiply that by 3 to get
30=9a+3b
Subtracting that from the third equation we find that
c=0
Subsituting that into the first two equations we find that
12=a+b 22=4a+2b
Doubling 12=a+b we find that
24=2a+2b
Subtracting that from 22=4a+2b we get that
-2=2a
and
-1=a
Which makes perfect sense seeing the parabola curves down
Subsituting this into 12=a+b we get
12=-1+b
and
13=b
Combining our results we find that
y=ax^2+bx+c
y=-1x^2+13x+0
y=-x^2+13x

Answer 4

Yes, the graph is indeed a parabola.

Its equation is y = -x^2 + 13x
Looking at your numbers one can see that the parabola is concave up and therefore the x^2 term must be negative.

Also it appears that the parabola is symmetric about the line x=6.5. The line of symmetry is given by y=-b/2a. Assuming a=-1, then -b/2 = 6.5, so b= 13

So for the parabola ax^2 + bx + c we have :
-x^2 + 13x +c =0
Put in any x,y pair and we find c=0.

Thus your equation is y = -x^2 + 13x

Hope this was clear.

Answer 5

So you're saying f(1)=12, f(2)=22, etc., right? (Just trying to understand the chart you drew above.)

For a parabola, we know the equation has to be second degree, so the equation is in some form of f(x) = ax² + bx + c

So f '(x) = 2ax + b, and since it appears that the vertex is at x=6.5, then we know that f '(6.5) = 13a+b = 0. {This is from 1st semester calculus; you didn't say what class you are in, so if you're not in calculus I apologize because what I just wrote won't make sense to you.}

Does this help you get started on the problem, or am I only making things worse for you?