Calculus question about tranformations and graphing.?
Alright, I'm just going to tell you how certain transformations affect graphs, and then you can do the rest. If we have y = f(x) then: y = f(x) + h Is a vertical shift "h" units. So y = f(x) + 1 moves the graph up one unit; negative h moves it down. ----------- y = f(x + k) Is a horizontal shift " -k" units. So y = f(x + 2) moves the graph LEFT 2 units, negative k moves it right. ----------- y = m f(x) Is a vertical stretch by a factor of m. So y = 4 f(x) multiplies every y value by 4. Note that if m = -1, we have a flip around the x axis. ----------- y = f(n x) Is a horizontal stretch by a factor of n. So y = f(3x) multiplies every x value by 3. Note that if n = -1, then we have a flip around the y axis. Done!
What is transformation ratio?
It is actually defined for a transformer.Transformation Ratio (K) is defined as the ratio of the EMF in the secondary coil to that in the primary coil.K = E2/E1 = (4.44(Φm)fN2)/(4.44(Φm)fN1)Therefore,K = E2/E1 = N2/N1….(1)Now,V1 = E1 + voltage dropE2 = V2 + voltage dropDue to the resistance in the windinds and some leakage flux, there is some loss in voltage. This is called as Voltage Drop.But, in ideal case, voltage drop can be neglected.Hence,V1 = E1E2 = V2Hence,E2/E1 = V2/V1…..(2)Also, in a transformer, the power across the primary as well as the secondary winding is same. Hence,V1.I1 = V2.I2V1/V2 = I2/I1……..(3)Now, combining (1), (2) & (3), we get,K = E2/E1 = N2/N1 = V2/V1 = I2/I1Where,1 represents the primary coil2 represents the secondary coilE is emf in the respective coilV is the voltage in the respective coilI is the current in the respective coilN is number of turns of the respective coilsΦm is the mutual flux in the core.
Graph the function and state the domain and range?
letting x - 2 = 0 you get that x = 2. This is the vertex of the "V" shaped graph that absolute values give you. When x = 2, y = 0 (2,0) Plot points on each side of x = 2, When x = 3, y = 1 (3,1) When x = 1, y= 1 (1,1) Draw the straight lines going from (0,2) through (3,1) and (0,2) through (1,1)---getting a V shape There are no domain restrictions on this function, so domain is (-infinity, infinity) . The vertex is the lowest point since the graph goes upwards, thus the lowest y value is y = 0 which makes the range= [0, infinity)
How can I transform continuous values into categorical variables in Python?
A simple function like pd.cut can transfer continuous values into categorical values.import pandas as pdimport numpy as np#generate a dataframe with random number for age from 0 - 9df_age = pd.DataFrame(np.random.randint(low=0, high=9, size=(100, 1)),columns = ['age'])#set up binsbin = [0,3,6,9]#use pd.cut function can attribute the values into its specific binscategory = pd.cut(df_age.age,bin)category = category.to_frame()category.columns = ['range']#concatenate age and its bindf_new = pd.concat([df_age,category],axis = 1)let’s draw a histogramimport matplotlib.pyplot as plt import seaborn as snssns.set(style="white")sns.set(style="whitegrid", color_codes=True)#draw histogram plotsns.countplot(x = 'range', data = df_new, palette = 'hls')plt.show()
What are the values of k for which the graph of y = x^3 - 3x^2 + k will have three distinct x-intercepts?
dy/dx = 3x^2 - 6x Stat Points: 3x(x-2) = 0 x = 0, 2 y = k, k-4 ie. stat pts at (0, k) and (2, k-4) If you draw the graph of a general cubic, you will see that the first stat pt must be above the x-axis, and the second one must be below, if there are to be three distinct x-intercepts. ie k > 0 AND k-4 < 0 (k<4) ie. 0 < k < 4
What are the values of k for which the graph of y = x^3 - 3x^2 + k will have three distinct x-intercepts?
A cubic function is x^3, so there has to be and will always be 3 zero's (some overlap, that's why they seem to only have only one or two). Whether they are all imaginary or real depends on the function(equation), but from there, we can see that y = x^3 - 3x^2 + k You want 3 DISTINCT zero's or x-ints: y = (x - a)(x - b)(x - c) y = (x^2 - ax - bx + ab)(x - c) = x^3 - cx^2 - ax^2 + acx - bx^2 + bcx + abx - abc Group them by x^3', x^2, x's and constants: = x^3 - ax^2 - bx^2 - cx^2 + abx + bcx + acx - abc = (1)x^3 + (-a - b - c)x^2 + (ab + bc + ac)x + (-abc) Compare this to your eqtn. y = x^3 - 3x^2 + k -a - b - c = -3 ab + bc + ac = 0 (since there's no x^1 term in your eqtn) -abc = k http://www.wolframalpha.com/input/?i=-a+-+b+-+c+%3D+-3%2C+ab+%2B+bc+%2B+ac+%3D+0%2C+-abc+%3D+k k = 0, 4 Therefore, 0 < k < 4 There's also a calculus way, but it includes taking derivatives and finding those points and plugging it back in to the original equation and see what it's between.
Function transformations help? - 10 points?
1. to graph the function g(x)= x^3-9 start with the graph f(x) = x^3 and then make which changes? shift the graph by what and which direction? reflect the graph over which axis? stretch the graph horizontally or vertically? (not all are necessary) and 2. to graph the function g(x)= 3sqrt(-x-10) start with the graph f(x) = 3sqrt(x) and then make which changes? shift the graph by what and which direction? reflect the graph over which axis? stretch the graph horizontally or vertically? (not all are necessary) THANK YOU!
How can I find the domain and range of a secant function?
Since the secant is the reciprocal of the cosine, it will not exist when the cosine x = 0. Thus, the domain is all x, except …-3π/2, -π/2, π/2, 3π/2, ……. The range will be the reciprocal of the range of the cosine function. Since the cosine has range [-1.1], the range of the secant is (-∞,-1] U [1,∞). If the range doesn’t make sense to you, take some values between -1 and 1 and flip them. You’ll see that the closer you get to zero for the cosine, the secant will get large without bound. For example, cos x could = 1/10000, so the corresponding secant would be 10000/1 or 10,000.