Maths-

General

Easy

Question

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript minus 7 semicolon a subscript 1 equals negative 3$

Hint:

A sequence is said to be arithmetic if the common difference is always constant.

The General formula of any AP is $a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d$ .

The correct answer is: a_n=4-7n.

Explanation:

We have given $a subscript n equals a subscript n minus 1 end subscript minus 7 comma a subscript 1 equals negative 3$

We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is $a subscript n equals a subscript n minus 1 end subscript plus d$ , where d is common difference.
Here we have $a subscript n equals a subscript n minus 1 end subscript minus 7$
So, d = -7
Also, We have a₁ = -3
So, The explicit formula will be
$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d end cell row cell a subscript n equals negative 3 plus left parenthesis n minus 1 right parenthesis left parenthesis negative 7 right parenthesis end cell row cell a subscript n equals 4 minus 7 n. end cell end table$
.

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript plus 1 semicolon a subscript 1 equals 12$

We have given $a subscript n equals a subscript n minus 1 end subscript plus 1 comma a subscript 1 equals 12$
We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here we have

a subscript n equals a subscript n minus 1 end subscript plus 1

So, d = 1
Also, We have a₁ = 12
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d end cell row cell a subscript n equals 12 plus left parenthesis n minus 1 right parenthesis left parenthesis 1 right parenthesis end cell row cell a subscript n equals 11 plus n end cell end table$

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript plus 1 semicolon a subscript 1 equals 12$

Maths-General

We have given $a subscript n equals a subscript n minus 1 end subscript plus 1 comma a subscript 1 equals 12$
We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here we have

a subscript n equals a subscript n minus 1 end subscript plus 1

So, d = 1
Also, We have a₁ = 12
So, The explicit formula will be

General

Maths-

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript minus 21 semicolon a subscript 1 equals 56$

We have given $a subscript n equals a subscript n minus 1 end subscript minus 21 comma a subscript 1 equals 56$
We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here we have

a subscript n equals a subscript n minus 1 end subscript minus 21

So, d = -21
Also, We have a₁ = 56
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d end cell row cell a subscript n equals 56 plus left parenthesis n minus 1 right parenthesis left parenthesis negative 21 right parenthesis end cell row cell a subscript n equals 77 minus 21 n end cell end table$

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript minus 21 semicolon a subscript 1 equals 56$

Maths-General

We have given $a subscript n equals a subscript n minus 1 end subscript minus 21 comma a subscript 1 equals 56$
We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here we have

a subscript n equals a subscript n minus 1 end subscript minus 21

So, d = -21
Also, We have a₁ = 56
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d end cell row cell a subscript n equals 56 plus left parenthesis n minus 1 right parenthesis left parenthesis negative 21 right parenthesis end cell row cell a subscript n equals 77 minus 21 n end cell end table$

General

Maths-

Graph each function $g left parenthesis x right parenthesis equals fraction numerator 3 x plus 2 over denominator x minus 1 end fraction$

Hint :-
Solution:-
1.Find the asymptotes of the rational function, if any.
2.Draw the asymptotes as dotted lines.
3.Find the x -intercept (s) and y -intercept of the rational function, if any.
4.Find the values of y for several different values of x .
5.Plot the points and draw a smooth curve to connect the points. Make sure that the graph does not cross the vertical asymptotes.
The vertical asymptote of a rational function is x -value where the denominator of the function is zero. Equate the denominator to zero and find the value of x .
x - 1= 0
x = 1
The vertical asymptote of the rational function is x= 1
We will find more points on the function and graph the function.

From the graph we can analyze that the vertical asymptote of the rational function is x = 1 and horizontal asymptote is
y = (leading coefficient of numerator) / (leading coefficient of denominator) =

3 over 1

Graph each function $g left parenthesis x right parenthesis equals fraction numerator 3 x plus 2 over denominator x minus 1 end fraction$

Maths-General

From the graph we can analyze that the vertical asymptote of the rational function is x = 1 and horizontal asymptote is
y = (leading coefficient of numerator) / (leading coefficient of denominator) =

3 over 1

General

Maths-

Find the equation for a path that passes through the point (6, 6) and is perpendicular to $y equals negative 3 over 4 x plus 1$ .

We have been given an equation that represents y-axis.
We have to find an equation for a path that passes through the point (6, 6) and is perpendicular to $y equals fraction numerator negative 3 over denominator 4 end fraction x plus 1$

Step 1 of 1:
We have given a line passes through a point (6, 6) and perpendicular to a line

y equals fraction numerator negative 3 over denominator 4 end fraction x plus 1

Since product of two perpendicular lines is equal to -1.
So,

$m cross times fraction numerator negative 3 over denominator 4 end fraction equals negative 1$

$m equals 4 over 3$
Therefore the equation of the line will be

$y minus 6 equals 4 over 3 left parenthesis x minus 6 right parenthesis$

$y equals 4 over 3 x minus 2$

Find the equation for a path that passes through the point (6, 6) and is perpendicular to $y equals negative 3 over 4 x plus 1$ .

Maths-General

We have been given an equation that represents y-axis.
We have to find an equation for a path that passes through the point (6, 6) and is perpendicular to $y equals fraction numerator negative 3 over denominator 4 end fraction x plus 1$

Step 1 of 1:
We have given a line passes through a point (6, 6) and perpendicular to a line

y equals fraction numerator negative 3 over denominator 4 end fraction x plus 1

Since product of two perpendicular lines is equal to -1.
So,

$m cross times fraction numerator negative 3 over denominator 4 end fraction equals negative 1$

$m equals 4 over 3$
Therefore the equation of the line will be

$y minus 6 equals 4 over 3 left parenthesis x minus 6 right parenthesis$

$y equals 4 over 3 x minus 2$

General

Maths-

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript minus 2 semicolon a subscript 1 equals negative 1$

We have given $a subscript n equals a subscript n minus 1 end subscript minus 2 comma a subscript 1 equals negative 1$
We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here we have

a subscript n equals a subscript n minus 1 end subscript minus 2

So, d = -2
Also, We have

a subscript 1 equals negative 1

So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d end cell row cell a subscript n equals negative 1 plus left parenthesis n minus 1 right parenthesis left parenthesis negative 2 right parenthesis end cell row cell a subscript n equals 1 minus 2 n end cell end table$

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript minus 2 semicolon a subscript 1 equals negative 1$

Maths-General

We have given $a subscript n equals a subscript n minus 1 end subscript minus 2 comma a subscript 1 equals negative 1$
We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here we have

a subscript n equals a subscript n minus 1 end subscript minus 2

So, d = -2
Also, We have

a subscript 1 equals negative 1

So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d end cell row cell a subscript n equals negative 1 plus left parenthesis n minus 1 right parenthesis left parenthesis negative 2 right parenthesis end cell row cell a subscript n equals 1 minus 2 n end cell end table$

General

Maths-

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript plus 6 semicolon a subscript 1 equals 9$

We have given $a subscript n equals a subscript n minus 1 end subscript plus 6 comma a subscript 1 equals 9$
We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here we have

a subscript n equals a subscript n minus 1 end subscript plus 6

So, d = 6
Also, We have a₁ = 9
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d end cell row cell a subscript n equals 9 plus left parenthesis n minus 1 right parenthesis 6 end cell row cell a subscript n equals 6 n plus 3 end cell end table$

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript plus 6 semicolon a subscript 1 equals 9$

Maths-General

We have given $a subscript n equals a subscript n minus 1 end subscript plus 6 comma a subscript 1 equals 9$
We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here we have

a subscript n equals a subscript n minus 1 end subscript plus 6

So, d = 6
Also, We have a₁ = 9
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d end cell row cell a subscript n equals 9 plus left parenthesis n minus 1 right parenthesis 6 end cell row cell a subscript n equals 6 n plus 3 end cell end table$

General

Maths-

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript plus 15 semicolon a subscript 1 equals 8$

We have given $a subscript n equals a subscript n minus 1 end subscript plus 15 comma a subscript 1 equals 8$
We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here we have

a subscript n equals a subscript n minus 1 end subscript plus 15

So, d = 15
Also, We have a₁ = 8
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d end cell row cell a subscript n equals 8 plus left parenthesis n minus 1 right parenthesis 15 end cell row cell a subscript n equals 15 n minus 7 end cell end table$

Write explicit formula. $a subscript n equals a subscript n minus 1 end subscript plus 15 semicolon a subscript 1 equals 8$

Maths-General

We have given $a subscript n equals a subscript n minus 1 end subscript plus 15 comma a subscript 1 equals 8$
We have to find the explicit formula of the given sequence.

Step 1 of 1:
We know that the recursive formula of an AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here we have

a subscript n equals a subscript n minus 1 end subscript plus 15

So, d = 15
Also, We have a₁ = 8
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript 1 plus left parenthesis n minus 1 right parenthesis d end cell row cell a subscript n equals 8 plus left parenthesis n minus 1 right parenthesis 15 end cell row cell a subscript n equals 15 n minus 7 end cell end table$

General

Maths-

Graph each function $f left parenthesis x right parenthesis equals fraction numerator 4 x minus 3 over denominator x plus 8 end fraction$

1.Find the asymptotes of the rational function, if any.
2.Draw the asymptotes as dotted lines.
3.Find the x -intercept (s) and y -intercept of the rational function, if any.
4.Find the values of y for several different values of x .
5.Plot the points and draw a smooth curve to connect the points. Make sure that the graph does not cross the vertical asymptotes.
The vertical asymptote of a rational function is x -value where the denominator of the function is zero. Equate the denominator to zero and find the value of x .
x + 8= 0
x = -8
The vertical asymptote of the rational function is x=-8
We will find more points on the function and graph the function.

From the graph we can analyze that the vertical asymptote of the rational function is x= -8 and horizontal asymptote is
y = (leading coefficient of numerator) / (leading coefficient of denominator) =

4 over 1

= 4

Graph each function $f left parenthesis x right parenthesis equals fraction numerator 4 x minus 3 over denominator x plus 8 end fraction$

Maths-General

From the graph we can analyze that the vertical asymptote of the rational function is x= -8 and horizontal asymptote is
y = (leading coefficient of numerator) / (leading coefficient of denominator) =

4 over 1

= 4

General

Maths-

What is the graph of the function $f left parenthesis x right parenthesis equals fraction numerator 2 x plus 1 over denominator 3 x minus 4 end fraction$

4 over 3

The vertical asymptote of the rational function is x=

4 over 3

We will find more points on the function and graph the function.

From the graph we can analyze that the vertical asymptote of the rational function is x=

4 over 3

. and horizontal asymptote is
y = (leading coefficient of numerator) / (leading coefficient of denominator) =

2 over 3

What is the graph of the function $f left parenthesis x right parenthesis equals fraction numerator 2 x plus 1 over denominator 3 x minus 4 end fraction$

Maths-General

4 over 3

The vertical asymptote of the rational function is x=

4 over 3

We will find more points on the function and graph the function.

From the graph we can analyze that the vertical asymptote of the rational function is x=

4 over 3

. and horizontal asymptote is
y = (leading coefficient of numerator) / (leading coefficient of denominator) =

2 over 3

General

Maths-

Write recursive formula and explicit formula. -15,-6,-3,12,21,...

We have given a sequence -15,-6,-3,12,21,.....
We have to find the recursive and explicit formula of the given sequence.

Step 1 of 2:
We have given a sequence -15,-6,-3,12,21,.....
The given sequence is an AP.
We know that the recursive formula for any AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here the common difference is 9
So, The recursive formula is

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript n minus 1 end subscript plus d end cell row cell a subscript n equals a subscript n minus 1 end subscript plus 9 end cell end table$
Step 2 of 2:
The given sequence is an AP.
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals negative 15 plus left parenthesis n minus 1 right parenthesis left parenthesis 9 right parenthesis end cell row cell a subscript n equals negative 15 plus 9 n minus 9 end cell row cell a subscript n equals 9 n minus 24 end cell end table$

Write recursive formula and explicit formula. -15,-6,-3,12,21,...

Maths-General

We have given a sequence -15,-6,-3,12,21,.....
We have to find the recursive and explicit formula of the given sequence.

Step 1 of 2:
We have given a sequence -15,-6,-3,12,21,.....
The given sequence is an AP.
We know that the recursive formula for any AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here the common difference is 9
So, The recursive formula is

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript n minus 1 end subscript plus d end cell row cell a subscript n equals a subscript n minus 1 end subscript plus 9 end cell end table$
Step 2 of 2:
The given sequence is an AP.
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals negative 15 plus left parenthesis n minus 1 right parenthesis left parenthesis 9 right parenthesis end cell row cell a subscript n equals negative 15 plus 9 n minus 9 end cell row cell a subscript n equals 9 n minus 24 end cell end table$

General

Maths-

The equation 2x + 7 represents a north path on a map. Find the equation for a path that passes through the point (6, 3) and is perpendicular to the north path.

We have been given an equation that represents the north path on a map.
We have to find an equation for a path that passes through the point (6, 3) and is perpendicular to the north path.

Step 1 of 1:
We have given a line passes through a point (6, 3) and perpendicular to a line

y equals 2 x plus 7

Since product of two perpendicular lines is equal to -1.
So,

$m cross times 2 equals negative 1$

$m equals fraction numerator negative 1 over denominator 2 end fraction$
Therefore the equation of the line will be

$y minus 3 equals fraction numerator negative 1 over denominator 2 end fraction left parenthesis x minus 6 right parenthesis$

$y equals fraction numerator negative 1 over denominator 2 end fraction x plus 6$

The equation 2x + 7 represents a north path on a map. Find the equation for a path that passes through the point (6, 3) and is perpendicular to the north path.

Maths-General

We have been given an equation that represents the north path on a map.
We have to find an equation for a path that passes through the point (6, 3) and is perpendicular to the north path.

Step 1 of 1:
We have given a line passes through a point (6, 3) and perpendicular to a line

y equals 2 x plus 7

Since product of two perpendicular lines is equal to -1.
So,

$m cross times 2 equals negative 1$

$m equals fraction numerator negative 1 over denominator 2 end fraction$
Therefore the equation of the line will be

$y minus 3 equals fraction numerator negative 1 over denominator 2 end fraction left parenthesis x minus 6 right parenthesis$

$y equals fraction numerator negative 1 over denominator 2 end fraction x plus 6$

General

Maths-

Write recursive formula and explicit formula. 62,57,52,47,42,...

We have given a sequence 62,57,52,47,42,....
We have to find the recursive and explicit formula of the given sequence.

Step 1 of 2:
We have given a sequence 62,57,52,47,42,....
The given sequence is an AP.
We know that the recursive formula for any AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here the common difference is -15
So, The recursive formula is

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript n minus 1 end subscript plus d end cell row cell a subscript n equals a subscript n minus 1 end subscript minus 15 end cell end table$
Step 2 of 2:
The given sequence is an AP.
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals 62 plus left parenthesis n minus 1 right parenthesis left parenthesis negative 15 right parenthesis end cell row cell a subscript n equals negative 15 n plus 77 end cell row cell a subscript n equals 77 minus 15 n end cell end table$

Write recursive formula and explicit formula. 62,57,52,47,42,...

Maths-General

We have given a sequence 62,57,52,47,42,....
We have to find the recursive and explicit formula of the given sequence.

Step 1 of 2:
We have given a sequence 62,57,52,47,42,....
The given sequence is an AP.
We know that the recursive formula for any AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here the common difference is -15
So, The recursive formula is

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals a subscript n minus 1 end subscript plus d end cell row cell a subscript n equals a subscript n minus 1 end subscript minus 15 end cell end table$
Step 2 of 2:
The given sequence is an AP.
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals 62 plus left parenthesis n minus 1 right parenthesis left parenthesis negative 15 right parenthesis end cell row cell a subscript n equals negative 15 n plus 77 end cell row cell a subscript n equals 77 minus 15 n end cell end table$

General

Maths-

Write recursive formula and explicit formula. -4, 5,14,23,32,...

Step 1 of 2:
We have given a sequence -4,5,14,23,32,...
The given sequence is an AP.
We know that the recursive formula for any AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here the common difference is 9
So, The recursive formula is

$a subscript n equals a subscript n minus 1 end subscript plus d$

$a subscript n equals a subscript n minus 1 end subscript plus 9$
Step 2 of 2:
The given sequence is an AP.
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals negative 4 plus left parenthesis n minus 1 right parenthesis 9 end cell row cell a subscript n equals 9 n minus 13 end cell row cell a subscript n equals 9 n minus 13 end cell end table$
.

Write recursive formula and explicit formula. -4, 5,14,23,32,...

Maths-General

Step 1 of 2:
We have given a sequence -4,5,14,23,32,...
The given sequence is an AP.
We know that the recursive formula for any AP is

a subscript n equals a subscript n minus 1 end subscript plus d

, where d is common difference.
Here the common difference is 9
So, The recursive formula is

$a subscript n equals a subscript n minus 1 end subscript plus d$

$a subscript n equals a subscript n minus 1 end subscript plus 9$
Step 2 of 2:
The given sequence is an AP.
So, The explicit formula will be

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a subscript n equals negative 4 plus left parenthesis n minus 1 right parenthesis 9 end cell row cell a subscript n equals 9 n minus 13 end cell row cell a subscript n equals 9 n minus 13 end cell end table$
.

General

Maths-

What are the vertical and horizontal asymptotes of the graph of each function?
$fraction numerator x squared plus 5 x plus 4 over denominator 3 x squared minus 12 end fraction$

1.Find the asymptotes of the rational function, if any.
2.Draw the asymptotes as dotted lines.
3.Find the x -intercept (s) and y -intercept of the rational function, if any.
4.Find the values of y for several different values of x .
5.Plot the points and draw a smooth curve to connect the points. Make sure that the graph does not cross the vertical asymptotes.
The vertical asymptote of a rational function is x -value where the denominator of the function is zero. Equate the denominator to zero and find the value of x .
3x² -12= 0
3x² = 12
x² = 4
x = -2 or x = 2
The vertical asymptote of the rational function is x = −2 and x = 2
We will find more points on the function and graph the function.

From the graph we can analyze that the vertical asymptote of the rational function is x= -2 and x = 2. and horizontal asymptote is
y = (leading coefficient of numerator) / (leading coefficient of denominator) =

1 third

What are the vertical and horizontal asymptotes of the graph of each function?
$fraction numerator x squared plus 5 x plus 4 over denominator 3 x squared minus 12 end fraction$

Maths-General

1.Find the asymptotes of the rational function, if any.
2.Draw the asymptotes as dotted lines.
3.Find the x -intercept (s) and y -intercept of the rational function, if any.
4.Find the values of y for several different values of x .
5.Plot the points and draw a smooth curve to connect the points. Make sure that the graph does not cross the vertical asymptotes.
The vertical asymptote of a rational function is x -value where the denominator of the function is zero. Equate the denominator to zero and find the value of x .
3x² -12= 0
3x² = 12
x² = 4
x = -2 or x = 2
The vertical asymptote of the rational function is x = −2 and x = 2
We will find more points on the function and graph the function.

1 third

General

Maths-

The equation 2x + 7 represents a north path on a map. Find the equation for a path that passes through the point (6, 3) and is parallel to the north path.

We have been given an equation that represents the north path on a map.
We have to find an equation for a path that passes through the point (6, 3) and is perpendicular to the north path.

Step 1 of 1:
We have given a line passes through a point (6, 3) and parallel to a line y = 2x + 7
Since two parallel lines have same slope.
So, Slope of the line will be 2
Therefore the equation of the line will be

y - 3 = 2(x - 6)
y = 2x - 9

The equation 2x + 7 represents a north path on a map. Find the equation for a path that passes through the point (6, 3) and is parallel to the north path.

Maths-General

We have been given an equation that represents the north path on a map.
We have to find an equation for a path that passes through the point (6, 3) and is perpendicular to the north path.

y - 3 = 2(x - 6)
y = 2x - 9

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

Write explicit formula.

A sequence is said to be arithmetic if the common difference is always constant. The General formula of any AP is .

The correct answer is: a_n=4-7n.

Explanation: We have given We have to find the explicit formula of the given sequence. Step 1 of 1:We know that the recursive formula of an AP is , where d is common difference.Here we have So, d = -7Also, We have a1 = -3So, The explicit formula will be .

Related Questions to study