Answer:
(-3 , 0) (0, 2) (6, -6)
Step-by-step explanation:
Solving for x - intercept:
2x - 3(0) = -6
2x = -6
2x/2 = -6/2
x = -3
One point would be (-3 , 0) .
Solving for y-intercept:
2(0) - 3y = -6
-3y = -6
-3y/-3 = -6/-3
y = 2
Another point would be (0, 2).
When 'x' equals 6:
2(6) - 3y = -6
12 - 3y = -6
12 - 12 - 3y = -6 - 12
-3y = -18
-3y/3 = -18/3
y = -6
The third point would be (6, -6).
Sas only because the triangles and usually congruent to me or in my opponion
Answer:
Step-by-step explanation:
B
Answer:
56
Step-by-step explanation:
Answer:
Analyzed and Sketched.
Step-by-step explanation:
We are given 
To sketch the graph we need to find 2 components.
1) First derivative of y with respect to x to determine the interval where function increases and decreases.
2) Second derivative of y with respect to x to determine the interval where function is concave up and concave down.

is absolute maximum

is the point concavity changes from down to up.
Here, x = 0 is vertical asymptote and y = 0 is horizontal asymptote.
The graph is given in the attachment.