Answer: GO TO MY ACCOUNT AND AWNSER MY QUESTION TEN I'LL AWNSER YOURS
Step-by-step explanation:
From what I know this graph looks like that of a modulus function except it's shifted 1 unit to right and 1 unit below the origin.
So the original equation must be:
y = |x|
And the transformed equation (aka the equation if this graph) must be,
y+1 = |x| -1
y = |x| -2
Answer:
B
Step-by-step explanation:
Because 15-4=11
and both angles are equal
There is a hole at x=4 because the top factors to (x-4)(x+3)
in the end we get
f(x)=(x-4)(x+3)/(x-4)
the (x-4) cancels out and we get
f(x)=x+3
so baiscally f(x) is y=x+3 but with a hole at (4,7)
one way is to make a piecewise function defined as(x^2-x-12)/(x-4) for all x≠4 and defined as 7 for x=4
or, to just do f(x)=x+3
![\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20)