The attached graph represents the graph of f(x) = (x - 1)^2 - 2
<h3>How to plot the graph?</h3>
The equation is given as:
f(x) = (x - 1)^2 - 2
Next, we set x to -2, -1, 0, 1 and 2.
So, we have:
f(-2) = (-2 - 1)^2 - 2 = 7
f(-1) = (-1 - 1)^2 - 2 = 2
f(0) = (0 - 1)^2 - 2 = -1
f(1) = (1 - 1)^2 - 2 = -2
f(2) = (2 - 1)^2 - 2 = -1
This means that the table of values is
x f(x)
-2 7
-1 2
0 -1
1 -2
2 -1
Next, we plot the above points and connect them.
See attachment for the graph of f(x) = (x - 1)^2 - 2
Read more about graphs and functions at:
brainly.com/question/4025726
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Answer:
an = a+4n-4
Step-by-step explanation:
Hello Friend,here is the solution for your question
<span>so the given function is </span>
y= √(-2cos²x+3cosx-1)
i.e = √[-2(cos²x-3/2+1/2)]
i.e = √[-2(cosx-3/4)²-9/16+1/2]
i.e. = √[-2(cos-3/4)²-1/16]
i.e. = √[1/8-3(cosx=3/4)²]-----------(1)
Now here in this equation is this quantity :-
<span>(cosx=3/4)²----------------(2) is to it's minimum value then the whole equation </span>
<span>i.e. = √[1/8-3(cosx=3/4)²] will be maximum and vice versa </span>
And we know that cosx-3/4 will be minimum if cosx=3/4
<span>therefore put this in (1) we get </span>
(cosx=3/4)²=0 [ cosx=3/4]
<span>hence the minimum value of the quantity (cosx=3/4)² is 0 </span>
<span>put this in equation (1) </span>
we get ,
i.e. = √[1/8-3(cosx=3/4)²]
=√[1/8-3(0)] [ because minimum value of of the quantity (cosx=3/4)² is 0 ]
=√1/8
=1/(2√2)
<span>this is the maximum value now to find the minimum value </span>
<span>since this is function of root so the value of y will always be ≥0 </span>
<span>hence the minimum value of the function y is 0 </span>
<span>Therefore, the range of function </span>y is [0,1/(2√2)]
__Well,I have explained explained each and every step,do tell me if you don't understand any step._
<u></u>
corresponds to TR. correct option b.
<u>Step-by-step explanation:</u>
In the given parallelogram or rectangle , we have a diagonal RT . We need to find which side is in correspondence with side/Diagonal RT of parallelogram URST .
<u>Side TU:</u>
In triangle UTR , we see that TR is hypotenuse and is the longest side among UR & TU . So , TR can never be equal in length to UR & TU . So there's no correspondence of Side TU with RT.
<u>Side TR:</u>
Since, direction of sides are not mentioned here , we can say that TR & RT is parallel & equal to each other . So , TR is in correspondence with side/Diagonal RT of parallelogram URST .
<u>Side UR:</u>
In triangle UTR , we see that TR is hypotenuse and is the longest side among UR & TU . So , TR can never be equal in length to UR & TU . So there's no correspondence of Side UR with RT.
The simple answer to your question is -36b-27
