Question

PLEASE HELP

Answer 1

Answer:

Step-by-step explanation:

You have 3 unknowns: a, b, and c.  It's our job to find them algebraically.  I'm going to start with the point where x = 0 and y = 7.  You'll see why in a minute.  Filling in the standard form of a quadratic

$y=ax^2+bx+c$  using (0, 7):

$7=a(0)^2+b(0)+c$  gives you that c = 7.  We will use that value now when we write the next 2 equations.  Now the point (-2, 19):

$19=a(-2)^2+b(-2)+7$  and

$19=4a-2b+7$ so

12 = 4a - 2b

Now for the next point (-1, 12):

$12=a(-1)^2+b(-1)+7$  and

$12=a-b+7$  so

5 = a - b

Now we have a system of equations (the 2 bold font equations) that we will solve by elimination:

 12  =  4a  -  2b

  5  =   a   -    b

Multiply the bottom equation by -4 to get a new system:

 12  =  4a  -  2b

-20  = -4a  +  4b

Add those together to get rid of the a terms and end up with

-8 = 2b so

b = -4

Now we can sub in -4 for b to solve for a.  I'm using the second bold type equation to do this:

5 = a - (-4) and

5 = a + 4 so

a = 1 and the equation for the quadratic function is

$y=x^2-4x+7$