Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
By using transformations sketch the graph of the given functions.
h ( x ) = ( x + 2)3
Solution
With these we have to first identify the "base" function. i.e. the function that's being shifted. In this case it seem like we are shifting f ( x ) = x3 . Then we can see that,
h ( x ) = ( x + 2)3 = f (x + 2)
In this case c = 2 and thus we're going to shift the graph of f ( x ) = x3 (the dotted line on the graph below) & move it 2 units to the left. It will mean subtracting two from the x coordinates of all the points on f ( x ) = x3 .
Following is the graph for this problem.
We have to probably do a quick review of intercepts before going much beyond. Intercepts are the points on which the graph will cross the x or y-axis. Determining intercepts is
graphing linear programs
linear functions
Solve A= P (1 + rt ) for r. Solution Here is an expression in the form, r = Equation involving numbers, A, P, and t In other terms, th
w2+30w+81 need step by step and explanation.
Next we desire to take a look at f (x ) =√x . First, note that as we don't desire to get complex numbers out of a function evaluation we ought to limit the values of x that we can
If a graph shows all of the possibilities for the no. of refrigerators and the no. of TVs that will fit into an 18 wheeler, will the truck hold 71 refrigerators and 118 Tvs?
a stack of disks is 0.5cm thick.how many disk each at 0.125cm thick are in a stack?
how to change the number to letter
how to add the positive numbers to negative numbers?and what is the highest +5 or -3
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +91-977-207-8620
Phone: +91-977-207-8620
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd