Examples of Rational Function Problems - Neurosparks (2023)

Solution

Step 1:We start by finding the intersections of the function:

• Öjsection is the point $latex(0,~ f(0))=(0, 3)$.
• ForX-Interceptions, we set the counter to zero and solve. Here, however, the numerator is the constant -3, so it has no zeros. Therefore the function does not haveX-intercepts.

Step 2:To find the vertical asymptotes, we set the denominator to zero and solve:

$latex x-1=0$

$latex x=1$

We have a vertical asymptote at $latex x=1$.

Level 3:The largest exponent ofXin the denominator is 1, which is greater than the exponent ofXin the counter (0). So theX-Axis is the horizontal asymptote.

Step 4:We only have one vertical asymptote, so we have two areas in the chart: $latex x>1$ and $latex x<1$.

We need a point in each region to determine whether it is above or below the horizontal asymptote. Therefore we can use:

$latex f(0)=3$ ⇒ $latex (0, ~3)$

$latex f(2)=-3$ ⇒ $latex (2, ~-3)$

Step 5:Here is the graph of the function:

Solution

Step 1:We need to find the sections of the function:

• Öj-intercept is the point $latex(0, ~f(0))=(0, ~4)$.
• we found themX-Sections by setting the counter to zero and solving:

$latex 4-2x=0$

$latex -2x=-4$

$latex x= 2$

ÖX-Intercept ist $latex x=2$.

Step 2:We find the vertical asymptotes by setting the denominator to zero and solving:

$latex 1-x=0$

$latex x=1$

We have a vertical asymptote at $latex x=1$.

Level 3:The greatest representatives ofXBoth the denominator and the numerator are the same. So the horizontal asymptote is equal to the coefficient ofXin the numerator divided by the coefficient ofXno denominator:

$latex y=\frac{-2}{-1}=2$

Step 4:We have a vertical asymptote, so we only have two areas in the chart: $latex x>1$ and $latex x<1$.

We need to find a point in each region to know if it's above or below the horizontal asymptote. So we use:

$latex f(0)=4$ ⇒ $latex (0, ~4)$

$latex f(2)=0$ ⇒ $latex (2, ~0)$

Step 5:With the obtained points we plot the function:

Solution

Step 1:We need to find the sections of the function:

• Öj-intercept is the point $latex(0, ~f(0))=(0, ~-2)$.
• we found themX-intercepts by setting the counter equal to zero and resolving. In this case, the numerator is the constant 4, so we don't have zeros and the function doesn'tX-intercepts.

Step 2:We set the denominator to zero and solve to find the vertical asymptotes:

$latex {{x}^2}+x-2=0$

$latex (x+2)(x-1)=0$

We have the vertical asymptotes $latex x=1$ and $latex x=-2$.

Level 3:In this function is the largest exponent ofXin the denominator is greater than the exponent ofXin the counter (0). So theX-Axis is the horizontal asymptote.

Step 4:We have two vertical asymptotes, i.e. three areas in the diagram: $latex x<-2$, $latex -2<x<1$ and $latex x>1$.

We need a point in each region to determine whether it is above or below the horizontal asymptote. The middle section is a bit trickier, so we need some points near the vertical asymptotes. Therefore we can use:

$latex f(-3)=1$ ⇒ $latex (-3, ~1)$

$latex f(-1)=-2$ ⇒ $latex (-1, ~-2)$

$latex f(0)=-2$ ⇒ $latex (0, ~-2)$

$latex f(2)=1$ ⇒ $latex (2, ~1)$

Step 5:Here is the graph of the function:

Solution

Step 1:The intersection points of the rational function are:

• If we use $latex x=0$ we get division by zero, so the function has no aj-intercept. However, we have already found a vertical asymptote.
• ÖX-Interruptions are:

$latex {{x}^2}-4=0$

$latex x=\pm 2$

So we have twoX-intercepts.

Step 2:We've already found one vertical asymptote, but there could be more. Therefore we set the denominator to zero and solve:

$latex {{x}^2}-4x=x(x-4)=0$

$latex x=0$ und $latex x=4$

We got two vertical asymptotes in $latex x=0$ and $latex x=4$.

Level 3:The largest exponent ofXin the numerator is 2, similar to the numerator. So we have a horizontal asymptote at:

$latex y=\frac{1}{1}=1$

Step 4:We have the regions: $latex x<0$, $latex 0<x<4$ and $latex x>4$.

One ofX-intercepts is in the left pane, so we don't need a dot there. The otherX-intercept is in the middle range, but we need more points to determine its behavior. Also, we need a point in the right region. So we have:

$latex f(1)=1$ ⇒ $latex (1,~1)$

$latex f(3)=-\frac{5}{3}$ ⇒ $latex (3, ~-\frac{5}{3})$

$latex f(5)=\frac{21}{5}$ ⇒ $latex (5, ~\frac{21}{5})$

Step 5:Here is the graph of the function:

Solution

Step 1:We start with the interceptions of the function:

• Öjsection is the point $latex(0, f(0))=(0, \frac{9}{2})$.
• ÖX-Interruptions are:

$latex 4{{x}^2}-36=0$

$latex 4{{x}^2}=36$

$latex {{x}^2}=9$

$latex x= \pm3$

ÖX-Sections are (-3, 0) and (3, 0).

Step 2:To find the vertical asymptotes, we set the denominator to zero and solve:

$latex {{x}^2}-2x-8=0$

$latex (x+2)(x-4)=0$

We have two vertical asymptotes in $latex x=-2$ and $latex x=4$.

Level 3:The largest exponent ofXin both the denominator and the numerator is 2. So the horizontal asymptote is:

$latex y= \frac{4}{1}=4$

Step 4:We have two vertical asymptotes, i.e. the regions: $latex x<-2$, $latex -2<x<4$ and $latex x>4$.

We need a point in each region, but we need some points in the middle region. Therefore we can use:

$latex f(-3)=0$⇒ $latex (-3, 0)$

$latex f(-1)= \frac{32}{5}$⇒ $latex (-1, \frac{32}{5})$

$latex f(3)=0$⇒ $latex (3, 0)$

$latex f(5)=\frac{64}{7}$⇒ $latex (5, \frac{64}{7})$

Step 5:Here is the graph of the function: