How do I find the horizontal asymptote of h(x)=x+6/x^2-64 ?
1 answer:
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The formula of linear equation is:
Where
x1 and x2: x coordinates(-1 and 3)
y1 and y2: y coordinates(-2 and 10)
m is the slope
We can then choose a point from the line to find the eqaution.
We need to first find the slope:
In this case:
In this case, as the y2 is given as 1,put (-1,-2) , and the slope(3 )into the eqaution :
y-(-2) = 3(x-(-1))
y+2 = 3(x+1)
Therefore
is the answer.
Hope it helps!
Where
x1 and x2: x coordinates(-1 and 3)
y1 and y2: y coordinates(-2 and 10)
m is the slope
We can then choose a point from the line to find the eqaution.
We need to first find the slope:
In this case:
In this case, as the y2 is given as 1,put (-1,-2) , and the slope(3 )into the eqaution :
y-(-2) = 3(x-(-1))
y+2 = 3(x+1)
Therefore
is the answer.
Hope it helps!
Answer:
The coordinates of J are missing. Plus, I don't see options for the conclusions. I made an imaginary point J, which you could correct to form the proper triangle.
Step-by-step explanation:
See the attachment. Plot all the vertices, including a corrected point J and draw the resulting triangle. It might be that AJKL is a slightly smaller, shifted version of AGHI. Enter the correct coordinates and then compare.
