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:

Step-by-step explanation:


given D : (7,-3), and D' : (2,5)
the coordinates of D can be represented as (x1,y1), and the coordinates of D' can be represented as (x,y).
you can simply take the difference in the x values and difference in the y values from the preimage to image.
like this:
f'(x,y) → f(x+(x-x1),y+(y-y1)) : 
D'(x,y) → D(x+(2-7),y+(5--3))
D'(x,y) → D(x<u>-5</u>,y<u>+8</u>) : 
Answer:
<em>In the next year, Anthony worked 2,084 hours</em>
Step-by-step explanation:
Anthony worked 1,697 hours in 2010.
We also know Anthony worked 22.8% more hours than in 2010.
The problem requires to calculate how much did Anthony work in the next year.
It can be calculated as follows:
Take 22.8% of 1,697:

Now calculate by adding it to the original number of hours:
1,697 + 387 = 2,084 hours
In the next year, Anthony worked 2,084 hours
Answer:
The horizontal shift is to the left by 2 units and the vertical shift is down 3 units for this function.
I hope I was able to help!
Step-by-step explanation: