Respuesta :
The correct answer is:
As x→-∞, y→-3.
As x→∞, y→∞.
Explanation:
As our values of x get further into the negative numbers, the value of 2ˣ will approach 0. This is because raising a number to a negative exponent "flips" the number below the denominator and raises it to a power; we end up with smaller and smaller fractions, eventually so small that they nearly equal 0.
This will make the value of the function 0-3=-3.
As x gets larger and larger (towards ∞), the value of y, 2ˣ, continues to grow as well. Since it continues to grow exponentially, we say the value approaches ∞.
As x→-∞, y→-3.
As x→∞, y→∞.
Explanation:
As our values of x get further into the negative numbers, the value of 2ˣ will approach 0. This is because raising a number to a negative exponent "flips" the number below the denominator and raises it to a power; we end up with smaller and smaller fractions, eventually so small that they nearly equal 0.
This will make the value of the function 0-3=-3.
As x gets larger and larger (towards ∞), the value of y, 2ˣ, continues to grow as well. Since it continues to grow exponentially, we say the value approaches ∞.
Answer:
For very high x-values, f(x) moves toward positive infinity.
For very low x-values, f(x) moves towards negative infinity.
Step-by-step explanation:
Here, the given function,
[tex]f(x) = 2x - 3[/tex]
Which is a polynomial function,
Since, the end behaviour of a polynomial depends upon the leading coefficient and degree of the polynomial,
If degree = odd and leading coefficient = positive,
Then end behaviour of the function f(x) is,
[tex]f(x)\rightarrow -\infty\text{ as }x\rightarrow -\infty[/tex]
[tex]f(x)\rightarrow +\infty\text{ as }x\rightarrow +\infty[/tex]
∵ Here, the degree is odd ( 1 ) and leading coefficient = 2 ( even )
Thus, the above end behaviour is also the end behaviour of the given function.
