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3 years ago

How do you find vertical, horizontal and oblique asymptotes for f(x) = (5x-15 )/ (2x 14)?

1 answer:

6 0

For vertical asymptotes, find the values which make the function indetermine in this case x=-7,so this is the only vertical asymptote.
For horizontal asymptotes, find the limit when x tends to infinity:
=(5x/x-15/x)/(2x/x+14/x) = 5/2, this is the horizontal asymptote y=5/2
For obliques, you have to meet the degree of the numerator is exactly a greater degree than the denominator, in this case they are the same degree so no oblique asymptote.

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b) 30 m/s

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The question is not complete, the correct question is given as:

The graph shows information about the speed of a vehicle during the final 50 seconds of a journey. At the start of the 50 seconds the speed is k metres per second. The distance travelled during the 50 seconds is 1.35 kilometres.

(a) Work out the average speed of the vehicle during the 50 seconds

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The graph is attached. The total time = 50 seconds, total distance = 1.35 km = 1350 m

a) The average speed is the ratio of the total distance traveled to the total time taken to cover this distance. The average speed is given by the formula:

$Average \ speed=\frac{total\ distance}{total\ time}\\ Substituting: \\ Average \ speed=\frac{1350\ m}{50\ s} = 27\ m/s$

b) From the graph, the total distance covered is the area of the graph. The graph is made up of a rectangle and triangle, the area of the graph is equal to the sum of area of rectangle and area of triangle.

$Total \ distance=Total\ area=Area\ of \ rectangle+Area\ of \ triangle\\Total \ distance=(length *breadth)+\frac{1}{2}base*height\\1350 \ m=(40*k)+0.5*10*k\\1350=40k+5k\\45k=1350\\k=1350/45\\k=30\ m/s$ c) i) The gradient in the last 10 seconds is the ratio of change in speed to change in time

$Gradient=\frac{change\ in\ speed }{change\ in\ time}=\frac{0-k}{50-40} \\ bt \ k=30\\Gradient=\frac{0-30}{50-40}= \frac{-30}{10} =-3$

ii) Since the gradient is negative it means it is deceleration. That is in the in the last 10 seconds the vehicle decelerates at a rate of 3 m/s²

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