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babunello [35]
3 years ago
10

In a rational function, is the horizontal shift represented by the vertical asymptote?

Mathematics
1 answer:
ziro4ka [17]3 years ago
7 0
<h3>Short Answer: Yes, the horizontal shift is represented by the vertical asymptote</h3>

A bit of further explanation:

The parent function is y = 1/x which is a hyperbola that has a vertical asymptote overlapping the y axis perfectly. Its vertical asymptote is x = 0 as we cannot divide by zero. If x = 0 then 1/0 is undefined.

Shifting the function h units to the right (h is some positive number), then we end up with 1/(x-h) and we see that x = h leads to the denominator being zero. So the vertical asymptote is x = h

For example, if we shifted the parent function 2 units to the right then we have 1/x turn into 1/(x-2). The vertical asymptote goes from x = 0 to x = 2. This shows how the vertical asymptote is very closely related to the horizontal shifting.

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Write an equation for the line described in standard form with x-intercept 4 and y-intercept 5
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Answer:

Equation( \frac{5}{4}) x  + y = 5 is in the standard dorm.

Step-by-step explanation:

Here, the x - intercept = 4 , so ( 4,0) is a point on the line.

and the y - intercept = 5, so ( 0,5) is a point on the line.

So, the slope of the equation is given as  =  \frac{y_2 - y_1}{x_2-x_1 }  = \frac{5 -0}{0-4}  = -\frac{5}{4}

Now, the SLOPE INTERCEPT FORM of an equation is :

y  = mx + b: here m  = slope and b =  y- intercept

or, y =-( \frac{5}{4}) x + 5

Now, standard form is Ax +By = C

So, the standard form is ( \frac{5}{4}) x  + y = 5

Hence, the above equation ( \frac{5}{4}) x  + y = 5

is in the standard dorm.

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Help!!! EASY!!! HELP DUE SOON!!!
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