Question

Hi, i need help with question 2! i need to graph all equations for precalculus!

Answer 1

To answer this question, we need to follow the same procedure. We need to find two points of the line using the given line equation.

Graph of the line y = 2x + 5

We have the slope-intercept form of the line here. We can find the x- and the y-intercepts to find two points of this line. The x-intercept is the point for x when y = 0. Likewise, the y-intercept is the point for y when x = 0.

Then, we have the x-intercept (the value of x when y = 0):

`0=2x+5\Rightarrow2x=-5\Rightarrow x=-(5)/(2)`

Then, the point is equal to (-5/2, 0).

The y-intercept is the value of y when x = 0:

`y=2\cdot(0)+5\Rightarrow y=5`

Therefore, the y-intercept is (0, 5).

Then, we have two points to graph this linear function: (-5/2, 0) and (0, 5).

We have that the graph is (for y = 2x + 5):

Graph of the line 2x + 3y = 6

This is the standard form of the line. We can find the x- and the y-intercept for this line following similar steps as the previous case:

x = 0 ---> to find the y-intercept of the line

`2\cdot(0)+3y=6\Rightarrow3y=6\Rightarrow(3y)/(3)=(6)/(3)\Rightarrow y=2`

Then, the y-intercept is (0, 2).

Now, we need to find the x-intercept (y = 0):

`2x+3(0)=6\Rightarrow2x=6\Rightarrow(2x)/(2)=(6)/(2)\Rightarrow x=3`

Then, the x-intercept is (3, 0).

Thus, the graph of the line is:

Graph of the equation y - 4 = 3(x+2)

We can rearrange the linear expression, and find the slope-intercept form of the line as follows:

`y-4=3x+6\Rightarrow y=3x+6+4\Rightarrow y=3x+10`

Now, we can follow the same steps to find the x- and the y-intercepts:

`y=0\Rightarrow0=3x+10\Rightarrow3x=-10\Rightarrow(3x)/(3)=-(10)/(3)\Rightarrow x=-(10)/(3)`

The x-intercept is (-10/3, 0).

`x=0\Rightarrow y=3(0)+10\Rightarrow y=10`

The y-intercept is (0, 10).

The graph for this line is:

Graph for the line 4(y - 5) = 2 (x + 1)

We can apply the distributive property to find the slope-intercept form of the line as we did with the above equations:

`4(y-5)=2(x+1)\Rightarrow4y-4\cdot5=2x+2\cdot1\Rightarrow4y-20=2x+2`

Then

`4y-20+20=2x+2+20\Rightarrow4y=2x+22\Rightarrow(4y)/(4)=(2x)/(4)+(22)/(4)`

Therefore

`y=(1)/(2)x+(11)/(2)`

And now, we can find the x- and the y-intercepts

`y=0\Rightarrow0=(1)/(2)x+(11)/(2)\Rightarrow2\cdot0=2\cdot((1)/(2)x+(11)/(2))\Rightarrow0=x+11\Rightarrow x=-11`

The x-intercept is (-11, 0)

The y-intercept is (x = 0):

`y=(1)/(2)x+(11)/(2)\Rightarrow y=(1)/(2)(0)+(11)/(2)\Rightarrow y=(11)/(2)`

Therefore, we have (0, 11/2), and the graph is:

Graph for the line y = x - 3 for x ∈ [-2, 6)

In this case, we need to graph for a restricted domain of the function. In this case, we have that the domain of the function is from -2 (inclusive) until 6 (but not equal to 6). Then, we can find the value of the function for x = -2, and x = 6, and draw a circle (not solid circle) for x = 6 (since the function is not defined for this value). Then, we have:

For x = -2

`y=x-3\Rightarrow y=-2-3\Rightarrow y=-5`

Then, the first point to graph the function is (-2, -5)

For x = 6

`y=x-3\Rightarrow y=6-3\Rightarrow y=3`

The other point is (6, 3).

Then, the graph of the function is:

We need to draw (carefully) a circumference for the point (6, 3) since the function is not defined for x = 6.

In summary, we have graphed the five lines finding the x- and the y-intercepts in the first four cases. In the last case, we found the value of the function for the extreme values of the domain for this linear function for x = -2, and for x = 6. Since the function is not defined for x = 6, we need to graph the linear function for a portion of the line since x = -2 until x = 6 (but not including the latter value). That is why the function is graphed as a small circumference for x = 6.