What are the points of discontinuity? Are they removable? y = (x-5) / x^2 - 6x + 5
1 answer:
Answer:
The points of discontinuity are: x=5 and x=1. The point of discontinuity x=5 can be removable.
Step-by-step explanation:
The points of discontinuity are those points where the function is not defined. To find such points, we should factorize the denominator of the function needs to be factorized.
Considering the quadratic equation of the form ax^2+bx+c =0, then using the quadratic (see attached image), where a=, b=- and c=5, we have that the roots are:
x=5 and x=1.
If we simplify the fraction, by removing the term (x-5) from both the numerator and the denominator, we get: , so we removed the point of discontinuity x=5.
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Given:
The parent function is
Consider the transformed function is g(x) instead of f(x) because both functions are different.
Step-by-step explanation:
We have,
It can be written as
...(i)
The translation is defined as
.... (ii)
Where, k is stretch factor, a is horizontal shift and b is vertical shift.
If 0<|k|<1, then the graph compressed vertically by factor k and if |k|>1, then the graph stretch vertically by factor k.
If k is negative, then f(x) is reflected across the x-axis to get g(x).
If a>0, then the graph shifts a units left and if a<0, then the graph shifts a units right.
If b>0, then the graph shifts b units up and if b<0, then the graph shifts b units down.
On comparing (i) and (ii), we get
, f(x) shifts 4 units right.
, f(x) shifts 8 units down.
, it is negative so f(x) reflected across the x-axis.
, so f(x) stretched vertically by factor 5.
Therefore, the function f(x) reflected across the x-axis, stretched vertically by factor 5 and shifted 4 units right 8 units down to get g(x).