Answer:
Function 1: None of the Above
Function 2: Quadratic
Function 3: Linear
Step-by-step explanation:
Function 1: It isn't linear because the y-axis doesn't go up or down at a constant rate, it isn't quadratic because it doesn't have a vertex, and it isn't exponential because it doesn't continue going up.
Function 2: It has a vertex-(6,32)
Function 3: The y-axis goes up at a constant rate of 6.
(Oya-hopes this helps.....)
Given:
Point is T(-3,8).
To find:
The coordinates of T' after
.
Solution:
We know that,
means the figure reflected across the x-axis then reflected across y-axis.
If a figure reflected across x-axis, then
![(x,y)\to (x,-y)](https://tex.z-dn.net/?f=%28x%2Cy%29%5Cto%20%28x%2C-y%29)
![T(-3,8)\to T_1(-3,-8)](https://tex.z-dn.net/?f=T%28-3%2C8%29%5Cto%20T_1%28-3%2C-8%29)
If a figure reflected across y-axis, then
![(x,y)\to (-x,y)](https://tex.z-dn.net/?f=%28x%2Cy%29%5Cto%20%28-x%2Cy%29)
![T_1(-3,-8)\to T'(-(-3),-8)](https://tex.z-dn.net/?f=T_1%28-3%2C-8%29%5Cto%20T%27%28-%28-3%29%2C-8%29)
![T_1(-3,-8)\to T'(3,-8)](https://tex.z-dn.net/?f=T_1%28-3%2C-8%29%5Cto%20T%27%283%2C-8%29)
Therefore, the required point is T'(3,-8).
Answer:
C
Step-by-step explanation:
I typed it into a graphing calculator. You could also choose a point (I choose (4,0) ) then plug it into each equation In the answers until you get back the numbers of the point you choose to represent x and y. For example
Yes it has the same value
They are exponents so it would be 8 times (4*4) + (2*2*2*2)
The exponent is how many times u multiple itself