What is the equation of the following graph in vertex form (0,12) (2,0) (4,-4) (6,0)

1 answer:

3 0

X value is going up and y value is moving up and down. So, it`s look like a quadratic function. y intercept is 12 and x intercept is 2 and 6. and vertex is (4,-4). than f(x)= a(x-4)^2+4. and put (0,12). and 16a+4=12. a=1/2. f(x)=1/2x^2-4x+12 or 1/2(x-4)^2+12

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Your task is to build a road joining a ranch to a highway that enables drivers to reach the city in the shortest time. The perpe

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Answer:

(a)In the attachment

(b)The road of length 35.79 km should be built such that it joins the highway at 19.52km from the perpendicular point P.

Step-by-step explanation:

(a)In the attachment

(b)The distance that enables the driver to reach the city in the shortest time is denoted by the Straight Line RM (from the Ranch to Point M)

First, let us determine length of line RM.

Using Pythagoras theorem

$|RM|^{2}=30^2+x^2\\|RM|=\sqrt{30^2+x^2}$

The Speed limit on the Road is 60 km/h and 110 km/h on the highway.

Time Taken = Distance/Time

Time taken on the road $=\frac{\sqrt{30^2+x^2}}{60}$

Time taken on the highway $=\frac{50-x}{110}$

Total time taken to travel, T $=\frac{\sqrt{30^2+x^2}}{60}+\frac{50-x}{110}$

Minimum time taken occurs when the derivative of T equals 0.

$T^{'}=\frac{x}{60\sqrt{30^2+x^2}}-\frac{1}{110}\\\frac{x}{60\sqrt{30^2+x^2}}-\frac{1}{110}=0\\\frac{x}{60\sqrt{30^2+x^2}}=\frac{1}{110}\\110x=60\sqrt{30^2+x^2}\\$

Square both sides

$12100x^2=3600(30^2+x^2)\\12100x^2=3240000+3600x^2\\12100x^2-3600x^2=3240000\\8500x^2=3240000\\x^2=\frac{3240000}{8500} =381.18\\x=\sqrt{381.18} =19.52$