Answer: Graph D will be correct graph for the given function.
Explanation:
Given function 
Since it is a bi-quadratic equation thus it must have 4 roots and (0,1) is one of its point.
Moreover, the degree of the function is even thus the end behavior of the function is
, as
and
as 
In graph A, function has four root but it does not have the end behavior same as function f(x).( because in this graph
, as
and
, as
.) so, it can not be the graph of given function.
In graph B, neither it has four root nor it has the end behavior same as function f(x).(because in this graph
as
and
as
.) so, it can not be the graph of given function.
In graph C, neither it has four root nor it has the same end behavior as function f(x).(because in this graph
as
and
as
.) so, it also can not be the graph of given function.
In graph D it has four root as well as it has the same end behavior as the given function. Also it passes through the point (0,1).
Thus, graph D is the graph of given function.
18 by 12
because 6*3=18 and then 4*3=12 so this is your answer
<span>1) Find P(E1UE2)
E1 probability= 1/2</span>
<span>There are 26 red cards in a 52 card deck, so the probability of choosing a red card is = 26/52 = 1/2
E2 probability= 12/ 52 or 3/13</span>
<span>The face cards are: Jacks, Queens, and <span>Kings. There are four suits, so in each suit there are one jack, one queen and one king. The probability is 3 x 4= 12 divided by the total number of cards.
2)</span></span><span>the probability of drawing a blue ball on the first draw: 4 /10
</span>the probability of drawing a white ball on the second drawn: 6/9 (because there is less one ball from the previous draw).
the probability of the cases together is 4/15 ( 4 /10 x 6/9) <span>since they are independent cases.</span>
Use the website: mathpapa.com/algebra-calculator.html
Answer:
Option C, the graph does not flip because 2 is positive