A function $f$ has a horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).
$ Part (a): Let $f$ be of the form $$f(x) = \frac{ax+b}{x+c}.$$Find an expression for $f(x).$
Part (b): Let $f$ be of the form $$f(x) = \frac{rx+s}{2x+t}.$$Find an expression for $f(x).$
Please explain
+981 What we need to do in this problem is to use the information that is given in the problem to find the expression for f(x). We are given the horizontal asymptote, the vertical asymptote, and the x-intercept.
To find the vertical asymptote, all you need to do is set the denominator equal to 0 and solve for x, but since we already know the vertical asymptote, we could work backwards to find the value of c.
Since the vertical asymptote is 3, we can plug in 3 for x and solve for c. 3+c=0
After solving this equation, we have c=-3
Our updated expression for f(x) becomes \(f(x)=\frac{ax+b}{x-3}\).
Finding the horizontal asymptote is different from finding the vertical asymptote.
There are three cases. Case 1: If the degree of numerator is less than the degree of the denominator, the horizontal asymptote is y=0.
Case 2: If the degree of numerator is equal to the degree of the denominator, you take the coefficients of the variable and you divide them.
Case 3: If the degree of the numerator is more than the degree of the denominator, there will be no horizontal asymptote. In this case, we see that the degree of the numerator and denominator is the same, they are both equal to 1, so we must divide the coefficients of the variable.
In the numerator, the coefficient of x is a, and in the denominator, the coefficient of x is 1.
So we can write the equation, \(a\div1=-4\)
After solving the equation, we have a=-4.
Now, our new updated expression is \(f(x)=\frac{-4x+b}{x-3}.\)
To find b, all we need to do is plug in the values for x and y the problem already gives us.
Plugging in 1 for x and 0 for y, we have: \(0=\frac{-4\cdot1+b}{1-3}\)
Solving the equation, we have $b=4.$ Now we have all four variables, the final expression for f(x) is: \(\boxed{f(x)=\frac{-4x+4}{x-3}} \)
For Part B, there are two methods.
Method 1: We can approach it the same way we did in Part A.
Once again, to find the vertical asymptote, we set our denominator equal to zero, and plug in 3 for x.
We have: \(2\cdot{3}+t=0\) Solving for t, we have t=-6
Also, the degree of the numerator is the same as the denominator, so you divide the coefficients of the x.
Since the horizontal asymptote is y=-4, we can write the equation: \(r\div2=-4.\)
Solving for r, we have r=-8
Our updated expression of f(x) is: \(f(x)=\frac{-8x+s}{2x-6}\).
Plugging in 1 for x and 0 for y, we have: \(\frac{-8\cdot{1}+s}{2\cdot{1}-6}=0\) Solving for s, we have s=8.
Our final expression for f(x) is \(\boxed{f(x)=\frac{-8x+8}{2x-6}}\)
Method 2: All we need to do to get the 2x in the denominator is multiply both the numerator and the denominator by 2 in the answer in Part A.
Our final expression for \(f(x) = \frac{2\cdot{-4}x+2\cdot{4}}{2\cdot{x}-2\cdot{3}}\) After simplifying, we have: \(\boxed{f(x)=\frac{-8x+8}{2x-6}}\)
I hope this helped,
Gavin.
+981 What we need to do in this problem is to use the information that is given in the problem to find the expression for f(x). We are given the horizontal asymptote, the vertical asymptote, and the x-intercept.
To find the vertical asymptote, all you need to do is set the denominator equal to 0 and solve for x, but since we already know the vertical asymptote, we could work backwards to find the value of c.
Since the vertical asymptote is 3, we can plug in 3 for x and solve for c. 3+c=0
After solving this equation, we have c=-3
Our updated expression for f(x) becomes \(f(x)=\frac{ax+b}{x-3}\).
Finding the horizontal asymptote is different from finding the vertical asymptote.
There are three cases. Case 1: If the degree of numerator is less than the degree of the denominator, the horizontal asymptote is y=0.
Case 2: If the degree of numerator is equal to the degree of the denominator, you take the coefficients of the variable and you divide them.
Case 3: If the degree of the numerator is more than the degree of the denominator, there will be no horizontal asymptote. In this case, we see that the degree of the numerator and denominator is the same, they are both equal to 1, so we must divide the coefficients of the variable.
In the numerator, the coefficient of x is a, and in the denominator, the coefficient of x is 1.
So we can write the equation, \(a\div1=-4\)
After solving the equation, we have a=-4.
Now, our new updated expression is \(f(x)=\frac{-4x+b}{x-3}.\)
To find b, all we need to do is plug in the values for x and y the problem already gives us.
Plugging in 1 for x and 0 for y, we have: \(0=\frac{-4\cdot1+b}{1-3}\)
Solving the equation, we have $b=4.$ Now we have all four variables, the final expression for f(x) is: \(\boxed{f(x)=\frac{-4x+4}{x-3}} \)
For Part B, there are two methods.
Method 1: We can approach it the same way we did in Part A.
Once again, to find the vertical asymptote, we set our denominator equal to zero, and plug in 3 for x.
We have: \(2\cdot{3}+t=0\) Solving for t, we have t=-6
Also, the degree of the numerator is the same as the denominator, so you divide the coefficients of the x.
Since the horizontal asymptote is y=-4, we can write the equation: \(r\div2=-4.\)
Solving for r, we have r=-8
Our updated expression of f(x) is: \(f(x)=\frac{-8x+s}{2x-6}\).
Plugging in 1 for x and 0 for y, we have: \(\frac{-8\cdot{1}+s}{2\cdot{1}-6}=0\) Solving for s, we have s=8.
Our final expression for f(x) is \(\boxed{f(x)=\frac{-8x+8}{2x-6}}\)
Method 2: All we need to do to get the 2x in the denominator is multiply both the numerator and the denominator by 2 in the answer in Part A.
Our final expression for \(f(x) = \frac{2\cdot{-4}x+2\cdot{4}}{2\cdot{x}-2\cdot{3}}\) After simplifying, we have: \(\boxed{f(x)=\frac{-8x+8}{2x-6}}\)
I hope this helped,
Gavin.
