Math / Algebra

Question $f(x) = \frac{3x+3}{x+2}$
Graph the rational function.
Start by drawing the vertical and horizontal asymptotes. Then plot two points on each piece of the graph. Finally, click on the graph-a-function button.

Studdy Solution

STEP 1

What is this asking? We need to graph the function $f(x) = \frac{3x+3}{x+2}$ , including asymptotes and two points on each piece of the graph. Watch out! Don't forget to consider both sides of the vertical asymptote when plotting points!

STEP 2

1. Find the vertical asymptote
2. Find the horizontal asymptote
3. Find points to the left of the vertical asymptote
4. Find points to the right of the vertical asymptote

STEP 3

A vertical asymptote occurs when the denominator of a rational function is equal to zero, and the numerator is not zero.
Let's set the denominator equal to zero and solve for $x$ : $x + 2 = 0$ $x = -2$ Since the numerator is $3(-2) + 3 = -3 \ne 0$ when $x = -2$ , we have a vertical asymptote at $x = -2$ .

STEP 4

To find the horizontal asymptote, we look at the degrees of the numerator and denominator.
Both the numerator, $3x + 3$ , and the denominator, $x + 2$ , have a degree of 1.
When the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients.

STEP 5

The leading coefficient of the numerator is 3, and the leading coefficient of the denominator is 1.
So, our horizontal asymptote is: $y = \frac{3}{1} = 3$

STEP 6

Let's choose $x = -3$ and $x = -4$ , which are to the left of our vertical asymptote at $x = -2$ .

STEP 7

For $x = -3$ , we have: $f(-3) = \frac{3(-3) + 3}{-3 + 2} = \frac{-9 + 3}{-1} = \frac{-6}{-1} = 6$ So, we have the point $(-3, 6)$ .

STEP 8

For $x = -4$ , we have: $f(-4) = \frac{3(-4) + 3}{-4 + 2} = \frac{-12 + 3}{-2} = \frac{-9}{-2} = \frac{9}{2} = 4.5$ So, we have the point $(-4, 4.5)$ .

STEP 9

Now let's choose $x = -1$ and $x = 0$ , which are to the right of our vertical asymptote at $x = -2$ .

STEP 10

For $x = -1$ , we have: $f(-1) = \frac{3(-1) + 3}{-1 + 2} = \frac{-3 + 3}{1} = \frac{0}{1} = 0$ So, we have the point $(-1, 0)$ .

STEP 11

For $x = 0$ , we have: $f(0) = \frac{3(0) + 3}{0 + 2} = \frac{0 + 3}{2} = \frac{3}{2} = 1.5$ So, we have the point $(0, 1.5)$ .

STEP 12

We found a vertical asymptote at $x = -2$ and a horizontal asymptote at $y = 3$ .
We also found two points on each side of the vertical asymptote: $(-3, 6)$ and $(-4, 4.5)$ to the left, and $(-1, 0)$ and $(0, 1.5)$ to the right.
Graph the asymptotes and plot these points to get the graph of $f(x)$ !

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Question $f(x) = \frac{3x+3}{x+2}$
Graph the rational function.
Start by drawing the vertical and horizontal asymptotes. Then plot two points on each piece of the graph. Finally, click on the graph-a-function button.

Studdy Solution

STEP 1

What is this asking? We need to graph the function $f(x) = \frac{3x+3}{x+2}$ , including asymptotes and two points on each piece of the graph. Watch out! Don't forget to consider both sides of the vertical asymptote when plotting points!

STEP 2

1. Find the vertical asymptote
2. Find the horizontal asymptote
3. Find points to the left of the vertical asymptote
4. Find points to the right of the vertical asymptote

STEP 3

STEP 4

To find the horizontal asymptote, we look at the degrees of the numerator and denominator.
Both the numerator, $3x + 3$ , and the denominator, $x + 2$ , have a degree of 1.
When the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients.

STEP 5

The leading coefficient of the numerator is 3, and the leading coefficient of the denominator is 1.
So, our horizontal asymptote is: $y = \frac{3}{1} = 3$

STEP 6

Let's choose $x = -3$ and $x = -4$ , which are to the left of our vertical asymptote at $x = -2$ .

STEP 7

For $x = -3$ , we have: $f(-3) = \frac{3(-3) + 3}{-3 + 2} = \frac{-9 + 3}{-1} = \frac{-6}{-1} = 6$ So, we have the point $(-3, 6)$ .

STEP 8

For $x = -4$ , we have: $f(-4) = \frac{3(-4) + 3}{-4 + 2} = \frac{-12 + 3}{-2} = \frac{-9}{-2} = \frac{9}{2} = 4.5$ So, we have the point $(-4, 4.5)$ .

STEP 9

Now let's choose $x = -1$ and $x = 0$ , which are to the right of our vertical asymptote at $x = -2$ .

STEP 10

For $x = -1$ , we have: $f(-1) = \frac{3(-1) + 3}{-1 + 2} = \frac{-3 + 3}{1} = \frac{0}{1} = 0$ So, we have the point $(-1, 0)$ .

STEP 11

For $x = 0$ , we have: $f(0) = \frac{3(0) + 3}{0 + 2} = \frac{0 + 3}{2} = \frac{3}{2} = 1.5$ So, we have the point $(0, 1.5)$ .

STEP 12

We found a vertical asymptote at $x = -2$ and a horizontal asymptote at $y = 3$ .
We also found two points on each side of the vertical asymptote: $(-3, 6)$ and $(-4, 4.5)$ to the left, and $(-1, 0)$ and $(0, 1.5)$ to the right.
Graph the asymptotes and plot these points to get the graph of $f(x)$ !

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Questionf(x)=3x+3x+2f(x) = \frac{3x+3}{x+2}f(x)=x+23x+3​Graph the rational function.Start by drawing the vertical and horizontal asymptotes. Then plot two points on each piece of the graph. Finally, click on the graph-a-function button.

Studdy Solution

STEP 1

What is this asking? We need to graph the function f(x)=3x+3x+2f(x) = \frac{3x+3}{x+2}f(x)=x+23x+3​, including asymptotes and two points on each piece of the graph. Watch out! Don't forget to consider both sides of the **vertical asymptote** when plotting points!

STEP 2

1. Find the vertical asymptote2. Find the horizontal asymptote3. Find points to the left of the vertical asymptote4. Find points to the right of the vertical asymptote

STEP 3

STEP 4

To find the **horizontal asymptote**, we look at the degrees of the numerator and denominator.Both the numerator, 3x+33x + 33x+3, and the denominator, x+2x + 2x+2, have a degree of **1**.When the degrees are the same, the **horizontal asymptote** is the ratio of the **leading coefficients**.

STEP 5

The **leading coefficient** of the numerator is **3**, and the **leading coefficient** of the denominator is **1**.So, our **horizontal asymptote** is: y=31=3y = \frac{3}{1} = 3y=13​=3

STEP 6

Let's choose x=−3x = -3x=−3 and x=−4x = -4x=−4, which are to the left of our **vertical asymptote** at x=−2x = -2x=−2.

STEP 7

For x=−3x = -3x=−3, we have: f(−3)=3(−3)+3−3+2=−9+3−1=−6−1=6f(-3) = \frac{3(-3) + 3}{-3 + 2} = \frac{-9 + 3}{-1} = \frac{-6}{-1} = 6f(−3)=−3+23(−3)+3​=−1−9+3​=−1−6​=6 So, we have the point (−3,6)(-3, 6)(−3,6).

STEP 8

For x=−4x = -4x=−4, we have: f(−4)=3(−4)+3−4+2=−12+3−2=−9−2=92=4.5f(-4) = \frac{3(-4) + 3}{-4 + 2} = \frac{-12 + 3}{-2} = \frac{-9}{-2} = \frac{9}{2} = 4.5f(−4)=−4+23(−4)+3​=−2−12+3​=−2−9​=29​=4.5 So, we have the point (−4,4.5)(-4, 4.5)(−4,4.5).

STEP 9

Now let's choose x=−1x = -1x=−1 and x=0x = 0x=0, which are to the right of our **vertical asymptote** at x=−2x = -2x=−2.

STEP 10

For x=−1x = -1x=−1, we have: f(−1)=3(−1)+3−1+2=−3+31=01=0f(-1) = \frac{3(-1) + 3}{-1 + 2} = \frac{-3 + 3}{1} = \frac{0}{1} = 0f(−1)=−1+23(−1)+3​=1−3+3​=10​=0 So, we have the point (−1,0)(-1, 0)(−1,0).

STEP 11

For x=0x = 0x=0, we have: f(0)=3(0)+30+2=0+32=32=1.5f(0) = \frac{3(0) + 3}{0 + 2} = \frac{0 + 3}{2} = \frac{3}{2} = 1.5f(0)=0+23(0)+3​=20+3​=23​=1.5 So, we have the point (0,1.5)(0, 1.5)(0,1.5).

STEP 12

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Question $f(x) = \frac{3x+3}{x+2}$
Graph the rational function.
Start by drawing the vertical and horizontal asymptotes. Then plot two points on each piece of the graph. Finally, click on the graph-a-function button.

What is this asking? We need to graph the function $f(x) = \frac{3x+3}{x+2}$ , including asymptotes and two points on each piece of the graph. Watch out! Don't forget to consider both sides of the vertical asymptote when plotting points!

1. Find the vertical asymptote
2. Find the horizontal asymptote
3. Find points to the left of the vertical asymptote
4. Find points to the right of the vertical asymptote

To find the horizontal asymptote, we look at the degrees of the numerator and denominator.
Both the numerator, $3x + 3$ , and the denominator, $x + 2$ , have a degree of 1.
When the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients.

The leading coefficient of the numerator is 3, and the leading coefficient of the denominator is 1.
So, our horizontal asymptote is: $y = \frac{3}{1} = 3$

Let's choose $x = -3$ and $x = -4$ , which are to the left of our vertical asymptote at $x = -2$ .

For $x = -3$ , we have: $f(-3) = \frac{3(-3) + 3}{-3 + 2} = \frac{-9 + 3}{-1} = \frac{-6}{-1} = 6$ So, we have the point $(-3, 6)$ .

For $x = -4$ , we have: $f(-4) = \frac{3(-4) + 3}{-4 + 2} = \frac{-12 + 3}{-2} = \frac{-9}{-2} = \frac{9}{2} = 4.5$ So, we have the point $(-4, 4.5)$ .

Now let's choose $x = -1$ and $x = 0$ , which are to the right of our vertical asymptote at $x = -2$ .

For $x = -1$ , we have: $f(-1) = \frac{3(-1) + 3}{-1 + 2} = \frac{-3 + 3}{1} = \frac{0}{1} = 0$ So, we have the point $(-1, 0)$ .

For $x = 0$ , we have: $f(0) = \frac{3(0) + 3}{0 + 2} = \frac{0 + 3}{2} = \frac{3}{2} = 1.5$ So, we have the point $(0, 1.5)$ .