Asked by Van
Graph the rational function
F(x)=6/-x-3
Draw the vertical and horizontal asymptotes then plot two points on each side of the graph
F(x)=6/-x-3
Draw the vertical and horizontal asymptotes then plot two points on each side of the graph
Answers
Answered by
Bot
To graph the rational function f(x) = 6/(-x - 3), we need to find the vertical and horizontal asymptotes first.
Vertical Asymptote:
The vertical asymptote occurs when the denominator is equal to zero.
Set -x - 3 = 0 and solve for x:
-x = 3
x = -3
So, the vertical asymptote is x = -3.
Horizontal Asymptote:
To find the horizontal asymptote, we examine the degree of the numerator and denominator.
The degree of the numerator is 0, and the degree of the denominator is 1.
Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is y = 0.
Now, let's plot two points on each side of the graph:
When x = -4:
f(-4) = 6/(-(-4) - 3) = 6/(-1) = -6
So, we have the point (-4, -6).
When x = -2:
f(-2) = 6/(-(-2) - 3) = 6/(-5) = -6/5
So, we have the point (-2, -6/5).
When x = -8:
f(-8) = 6/(-(-8) - 3) = 6/5
So, we have the point (-8, 6/5).
When x = -6:
f(-6) = 6/(-(-6) - 3) = -6/5
So, we have the point (-6, -6/5).
Here is the graph of the rational function with the vertical and horizontal asymptotes, as well as the plotted points:
{{{ graph( 300, 300, -10, 10, -10, 10, 6/(-x - 3)) }}}
Note: The graph has been shifted to the left by 3 units due to the negative sign in front of the x in the denominator.
Vertical Asymptote:
The vertical asymptote occurs when the denominator is equal to zero.
Set -x - 3 = 0 and solve for x:
-x = 3
x = -3
So, the vertical asymptote is x = -3.
Horizontal Asymptote:
To find the horizontal asymptote, we examine the degree of the numerator and denominator.
The degree of the numerator is 0, and the degree of the denominator is 1.
Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is y = 0.
Now, let's plot two points on each side of the graph:
When x = -4:
f(-4) = 6/(-(-4) - 3) = 6/(-1) = -6
So, we have the point (-4, -6).
When x = -2:
f(-2) = 6/(-(-2) - 3) = 6/(-5) = -6/5
So, we have the point (-2, -6/5).
When x = -8:
f(-8) = 6/(-(-8) - 3) = 6/5
So, we have the point (-8, 6/5).
When x = -6:
f(-6) = 6/(-(-6) - 3) = -6/5
So, we have the point (-6, -6/5).
Here is the graph of the rational function with the vertical and horizontal asymptotes, as well as the plotted points:
{{{ graph( 300, 300, -10, 10, -10, 10, 6/(-x - 3)) }}}
Note: The graph has been shifted to the left by 3 units due to the negative sign in front of the x in the denominator.
Answered by
Bot
Using the properties of logarithms, we can expand log_z(x^4) as follows:
log_z(x^4) = 4log_z(x)
So, log_z(x^4) expands to 4log_z(x).
log_z(x^4) = 4log_z(x)
So, log_z(x^4) expands to 4log_z(x).
Answered by
Bot
To solve for x, we need to rewrite the equation using exponential form.
In logarithmic form, the equation log base 10 x = -1 means that 10 raised to the power of -1 is equal to x.
Thus, x = 10^(-1).
Evaluating 10^(-1), we find that x = 1/10 or 0.1.
In logarithmic form, the equation log base 10 x = -1 means that 10 raised to the power of -1 is equal to x.
Thus, x = 10^(-1).
Evaluating 10^(-1), we find that x = 1/10 or 0.1.
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