Answer:
[tex]y=-3x^2+11x-6[/tex]
Step-by-step explanation:
Recall that one needs just three points on the plane to determine the form of the quadratic function that goes through them, because a quadratic function is defined by three parameters. We therefore can pick just three simple ones that facilitate our calculations.
Realize that we need to find the parameters [tex]a,\,b,[/tex] and [tex]c[/tex] that gives as a function of the form: [tex]y=ax^2+bx+c[/tex] that satisfies the values given in the table.
The simplest point to start with is the point (0,-6) which means that when x is 0 (zero), the value of y must be "-6". Placing such values in the general form, renders what the value of the parameter "c" should be:
[tex]y=ax^2+bx+c\\-6=a(0)^2+b(0)+c\\c=-6[/tex]
So parameter [tex]c[/tex] must be "-6".
Now let's use other simple values of "x" that facilitate our calculations, and include the value we just found, to reduce the number of unknowns:
point (1, 2):
[tex]y=ax^2+bx-6\\2=a(1)^2+b(1)-6\\8=a+b[/tex]
Point (-1, -20):
[tex]y=ax^2+bx-6\\-20=a(-1)^2+b(-1)-6\\-14=a-b[/tex]
Now, we can add these two equations term by term as they are, in order to get rid of the unknown "b":
[tex]a+b=8\\a-b=-14\\------\\2a+0=-6\\2a=-6\\a=\frac{-6}{2} \\a=-3[/tex]
So we just found the value for parameter "[tex]a[/tex]" = -3
Now we can use it as well as c = -6 in one of the equations we reduced above, in order to find the parameter (b) that we are missing:
[tex]a+b=8\\-3+b=8\\b=8+3\\b=11[/tex]
So now we have the three parameters we need to define the quadratic function: [tex]y=-3x^2+11x-6[/tex]
You can also check that the other points given in the table (2, 4) a,d (3, 0) are indeed points that belong to this quadratic function.
