B. 36
<span>Multiply or count branches it's up to you.</span>
Answer:
f(x) = 3x+2
g(x) = 2x - 2
To find (f-g)(x)
(f-g)(x) = f(x) - g(x)
Substitute f(x) to 3x+2 and g(x) to 2x -2
(f-g)(x) = (3x+2) - (2x-2)
Open the brackets
(f-g)(x) = 3x+2-2x+2
Rearrange
(f-g)(x) = 3x-2x+2+2
(f-g)(x) = x + 4
Answer is option A = x + 4
Step-by-step explanation:
Answer:
your answer would be C
Step-by-step explanation:
line segment JK,Modifying above upper L upper J with bar, Modifying above Upper L Upper K with bar
<span>Vector Equation
(Line)</span>(x,y) = (x,y) + t(a,b);tERParametric Formx = x + t(a), y = y + t(b); tERr = (-4,-2) + t((-3,5);tERFind the vector equation of the line passing through A(-4,-2) & parallel to m = (-3,5)<span>Point: (2,5)
Create a direction vector: AB = (-1 - 2, 4 - 5)
= (-3,-1) or (3,1)when -1 (or any scalar multiple) is divided out.
r = (2,5) + t(-3,-1);tER</span>Find the vector equation of the line passing through A(2,5) & B(-1,4)<span>x = 4 - 3t
y = -2 + 5t
;tER</span>Write the parametric equations of the line passing through the line passing through the point A(4,-2) & with a direction vector of m =(-3,5)<span>Create Vector Equation first:
AB = (2,8)
Point: (4,-3)
r = (4,-3) + (2,8); tER
x = 4 + 2t
y = -3 + 8t
;tER</span>Write the parametric equations of the line through A(4,-3) & B(6,5)<span>Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in -3
-3 = 5 + 4t
(-8 - 5)/4 = t
-2 = t
For y sub in -8
-8 = -2 + 3t
(-8 + 2)/3 = t
-2 = t
Parameter 't' is consistent so pt(-3,-8) is on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (-3,-8) on the line?<span>Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in 1
-1 = 5 + 4t
(-1 - 5)/4 = t
-1 = t
For y sub in -7
-7 = -2 + 3t
(-7 + 2)/3 = t
-5/3 = t
Parameter 't' is inconsistent so pt(1,-7) is not on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (1,-7) on the line?<span>Use parametric equations when generating points:
x = 5 + 4t
y = -2 + 3t ;tER
X-int:
sub in y = 0
0 = -2 + 3t
solve for t
2/3 = t (this is the parameter that will generate the x-int)
Sub t = 2/3 into x = 5 + 4t
x = 5 + 4(2/3)
x = 5 + (8/3)
x = 15/3 + (8/3)
x = 23/3
The x-int is (23/3, 0)</span>What is the x-int of the line r = (5,-2) + t(4,3); tER?Note: if they define the same line: 1) Are their direction vectors scalar multiples? 2) Check the point of one equation in the other equation (LS = RS if point is subbed in)What are the two requirements for 2 lines to define the same line?
Start from the parent function 
In the first case, you are computing
In the second case, you are computing
![f(x+1)-2 /tex] There are two translation going on: when you transform [tex] f(x) \to f(x+k)](https://tex.z-dn.net/?f=%20f%28x%2B1%29-2%20%2Ftex%5D%3C%2Fp%3E%20%3Cp%3EThere%20are%20two%20translation%20going%20on%3A%20when%20you%20transform%20%5Btex%5D%20f%28x%29%20%5Cto%20f%28x%2Bk%29%20)
, you translate the function horizontally, 
units left if 
and units right if 
.
On the other hand, when you transform 
, you translate the function vertically, units up if and units down if .
So, the first function is the "original" parabola , translated 
units right and 
units up. Likewise, the second function is the "original" parabola , translated 
units left and 
units down.
So, the transformation from 
to 
is: go 
units to the left and units down
