First set up a linear equation and using the x and y values in the table see if it solves.
It doesn't solve so we know it isn't linear. ( I won't show all those steps because they aren't needed.)
Using the quadratic formula y = ax^2 +bx +c
Build a set of 3 equations from the table:
C is the Y intercept ( when X is 0), this is shown in the table as 6
Now we have y = ax^2 + bx + 6
-2.4 =4a-2b +6
1.4 = a-b +6
Rewrite the equations
a=b/2 -2.1
1.4 = b/2-2.1 +6
b = 5
a = 5/2 -2.1 = 0.4
replace the letters to get y = 0.4x^2 + 5x +6
Answer:
The coordinates of point c: (7,5)
The coordinates of point D:(8,1)
Um I think 4 i hope this helps
Answer:
2009000
Step-by-step explanation:
Answer:
Step-by-step explanation:
Take the derivate of g:
Find x that:
g'(x)=0
<h2>solving:</h2>
This x give the least possible value that are g(7/2):
