See the attached image for the graph. Specifically figure 2 is the graph you want. You can leave the red points on the graph or decide to erase them (leave behind the blue line though).
To generate each of the red points, you'll plug in various x values to get corresponding y values.
For instance, plug in x = 0 and we get...
y = -|x-6| - 6
y = -|0-6| - 6
y = -|-6| - 6
y = -6 - 6
y = -12
So when x = 0, the y value is -12. The x and y value pair up to get (x,y) = (0,-12)
Another example: plug in x = 2
y = -|x-6| - 6
y = -|2-6| - 6
y = -|-4| - 6
y = -4 - 6
y = -10
So the point (2,-10) is on the graph
The idea is to generate as many points as possible so we get an idea of what this thing looks like.
Generate enough points, and you'll get what you see in Figure 1 (see attached image)
Then draw a line through all of the points. The more points you use, the more accurate the drawing. Doing that will generate the blue function curve you see in Figure 2 (also attached)
Answer:
a- x = 5/3, or x = -7/2
b- 675
c - 5·x + 2
Step-by-step explanation:
The polynomial representing the capital of the two partners = 6·x² + 11·x - 35
a. The total share is the capital of the two partners together = 6·x² + 11·x - 35
∴ When their total share is equal to 0, we have;
6·x² + 11·x - 35 = 0
Factorizing the above equation with a graphing calculator gives;
(3·x - 5)·(2·x + 7)
Therefore;
x = 5/3, or x = -7/2
b- The total expenditure, when x = 10 is given by substituting the value of <em>x </em>in the polynomial 6·x² + 11·x - 35, as follows;
When x = 10
6·x² + 11·x - 35 = 6 × 10² + 11 × 10 - 35 = 675
The total expenditure of Vicky and Micky when x = 10 is 675
c - The sum of their expenditure is (3·x - 5) + (2·x + 7) = 5·x + 2
Answer:
y=4
Step-by-step explanation:
From the graph we can see that-
when, x=-3 , y=4
That is , the graph passes through (-3 , 4)
So,
for x=-3 ,
we get y=4
So, the answer is y=4
ANSWER



EXPLANATION
The given quadratic equation is:

We rewrite in the standard quadratic equation form to obtain,

Comparing this to the general standard quadratic equation.

We have my


