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Marina86 [1]
3 years ago
12

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

Mathematics
1 answer:
Andrei [34K]3 years ago
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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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

(b) Work out the value of k.

(c) (i) Calculate the gradient of the graph in the final 10 seconds of the journey

    (ii) Describe what this gradient represents

Answer:

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/sc) 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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