Determining a 2nd degree function

• We know that a quadratic function is an expression f(x)=ax2+bx+c.

  • For example, f(x)=3x2-5x+7

• When we input a value into the variable x, we find its image by performing the calculation.

  • For instance, in the function f(x)=3x2-5x+7, if we plug in the value x=2, we obtain the image f(2)=3·22-5·2+7=12-10+7=9

• But what happens if we only know that it is a quadratic function, but we don't know its coefficients?

• That is, if we only know that it is a function of the form f(x)=ax2+bx+c, but we don't know the values of a, b, and c?

• In this case, if we know some images of the function, we can determine what the coefficients of f(x) are by using that information.

We know that the function f(x) is quadratic, and its graph passes through the points (0,4), (-2,18), and (3,13).

(1) We know that f(x)=ax2+bx+c

(2) If we plug in x=0, the result must be 4 --> f(0)=a·02+b·0+c=4 --> c=4

(3) If we plug in x=-2, the result must be 18 --> f(-2)=a·(-2)2+b·(-2)+c=18 --> 4a-2b+c=18

(4) If we plug in x=3, the result must be 13 --> f(3)=a·32+b·3+c=13 --> 9a+3b+c=13

(5) We have, therefore, 3 equations:

• c = 4

• 4a-2b+c = 18

• 9a+3b+c = 13

(6) From the 1st equation, we know that c = 4, and we can plug it into the other two equations:

• 4a-2b+4 = 18

• 9a+3b+4 = 13

(7) Therefore...

• 4a - 2b = 14

• 9a + 3b = 9

(8) We solve the system, either by substitution, equalization or elimination, and we find out:

• a = 2

• b = -3

(9) Therefore, the function is:

• • f(x) = 2x2-3x+4

Follow the previous instructions to find out the expressions of a quadratic function f(x) whose graph passes through the points (0,5), (1,6), and (-2,-3).