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i need on these and it doesnt really matter but theres alot of points w/ answering!
1. use the graph below to plot the line shown by the slope of 4/3 and a y intercept of -3. explain how you were able to create the graph from this information and then find the equation of that line in slope intercept form
2. graph the linear inequality 6x + 2y > 4. you will need to show the solution area by shading the appropriate side of the line. show all of your work on how you determined where to shade for the solution set.
3. using the equation below:
9x – 3y = 12
a) take the equation and change it into slope intercept form.
b) determine the y intercept and slope of this linear equation.
c) find the equation of the line that passes through the point, ( 3, 5 ) and is perpendicular to the original line.

Step-by-step explanation:

1. We are given that slope is 4/3 and y-intercept is -3.

Now, the general form of a straight line is y = mx + c where m is the slope and c is the y-intercept.

So, in our case, the equation of the line is .

Now, substituting x=0 and y=0, we get the pair of points (x,y) = (0,-3) , (,0).

So, plotting these points on a graph and joining them gives the required line as seen in graph 1 below.

We have, the slope intercept form of the line is .

2. We have the inequality 6x + 2y > 4.

In order to find the solution area, we will use the 'Zero Test' i.e. substitute x=0 and y=0. If the inequality is true, the shaded area is towards the origin and if its false, the shaded area is away from the origin.

Now, we have 6x + 2y > 4 → 0 > 4 which is false. So, the solution region is away from the origin as seen in the graph 2 below.

3. We have the equation 9x - 3y = 12.

A) The slope intercept form is y = 3x - 4.

Because, 9x - 3y = 12 →  3y = 9x - 12 → y = 3x - 4.

B) Now comparing y = 3x - 4 by the general form of a straight line, we see that slope of this line is 3 and y-intercept is -4.

C) We have to find the equation of line perpendicular to y = 3x - 4 having slope m_{1} = 3 and passing through ( 3,5 ).

As the lines are perpendicular, the slopes have the relation,

i.e.

i.e.

Now, using this slope and the point ( 3,5 ) , we will find the equation of the line using the formula,

i.e.

i.e.

i.e.

So, the line perpendicular to y = 3x - 4 having slope m_{1} = 3 and passing through ( 3,5 ) is .

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.