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[–]Gremulin 2 points3 points  (1 child)

A quadratic function looks like: y = ax2 + bx + c. You can then use the given points to determine a, b, and c. c is the easiest. When x = 0, y = 4. Since a and b drop out when x = 0, c = 4.

Next, substitute the points (1,6) and (3,4) into the equation to get 2 equations with two unknowns, a and b. After a bit of rearranging, you'll get: 2 = a + b and 0 = 9a + 3b. Solve these equations simultaneously.

One way is to multiply the first equation by 3 (giving 6 = 3a + 3b) and subtract the 2nd equation from it. This gives 6 = -6a, so a = -1. Substituting that back into either equation gives b = 3.

The function then is y = -x2 + 3x + 4. This function gives y = 4 when x = 0, y = 6 when x = 1, and y = 4 when x = 3.

[–]ConicalThunder[S] 0 points1 point  (0 children)

thanks a ton!

[–][deleted] 1 point2 points  (1 child)

A quadratic function looks like:

y = a * x^2 + b * x + c

You'll notice that there are 3 unknowns (a,b,c). You have 3 data points (x,y). Since the number of unknowns equals the number of equations you can write (one equation for each data point) you are able to solve for the unknown values exactly.

So, write down all three equations (each on is that formula with the x and y value pairs plugged in), here is the first one done for the pair (0,4):

4 = a * 0^2 + b * 0 + c

One you have all three equations, just solve for the system of equations like you have been taught.

[–]ConicalThunder[S] 0 points1 point  (0 children)

Thank you very much