Write the quadratic function the points are (-1,35) (0,22) (1,11) (2,2) and (3,-5)

Step-by-step explanation:

We only need 3 of those points and the standard form of the quadratic to find the function.  The standard form is

From each coordinate we will sub in the x and y values.  I am going to use the following 3 points from the table. (0, 22), (1, 11) and (2, 2).  Always start with a coordinate that has an x value of 0 if you can!  Filling in the standard form with the first of those 3 coordinates:

simplifies down to the fact that c = 22. That's good to know; now we have a value for one of the variables to fill into the next substitution:

which simplifies down to

a + b = -11

Now for the 3rd and last coordinate:

which simplifies down to

4a + 2b = -20

Put the 2 bold equations together as a system and solve for one variable or the other.  I chose to solve for b first, so in order to do that, I multiplied the first equation by a -4, giving me a new system:

 -4a  -  4b  =  44

  4a  +  2b  = -20

and

-2b = 24 so

b = -12. Now plug that back in and solve for a:

a - 12 = -11 so

a = 1.

Now we have the function:

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.

1. 16.64

a^2+b^2=c^2

pr^2=pq^2+qr^2

pr^2=14^2+ 9^2

pr^2=196+81

pr= sqr root of 277

16.64

2. 11.31

c^2-a^2=b^2

18^2- 14^2=b^2

b^2=324-196

b=sqr root of 126

b=11.31

hope it ! (btw im in connections academy to, good luck! )