Question

Type the correct answer in the box.

Answer 1

Answer:

$y=-3x^2+11x-6$

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 $a,\,b,$ and $c$ that gives as a function of the form: $y=ax^2+bx+c$ 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:

$y=ax^2+bx+c\\-6=a(0)^2+b(0)+c\\c=-6$

So parameter $c$ 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):

$y=ax^2+bx-6\\2=a(1)^2+b(1)-6\\8=a+b$

Point (-1, -20):

$y=ax^2+bx-6\\-20=a(-1)^2+b(-1)-6\\-14=a-b$

Now, we can add these two equations term by term as they are, in order to get rid of the unknown "b":

$a+b=8\\a-b=-14\\------\\2a+0=-6\\2a=-6\\a=\frac{-6}{2} \\a=-3$

So we just found the value for parameter "$a$" = -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:

$a+b=8\\-3+b=8\\b=8+3\\b=11$

So now we have the three parameters we need to define the quadratic function: $y=-3x^2+11x-6$

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.