unit 2 methods help

[sally baker]

how can you find the stationary points and coordinates of this


y=(x+1)(x+4)
i did it and im getting x=3/2 however the answer says x=-3



and for this question y=ax^2+bx has a gradient of 3 at point (2,-2)
how do you find coordinates of turning points



Also
i find it difficult finding the average velocity in 5 seconds & average speed in 5 seconds
 e.g for the equation   s=t^3-3t^2+2t

Also for the equation 5/8(10t^2-t^3/3 ) ( 20 is less than or equal to t , as t is less then or equal to 0)





thanks

[sally baker]

for the last equation  V= 5/8(10t^2-t^3/3 )  i  need to find its turning point the books answer says  (10,62.5) as turning points , and the t values are 0 and 20 as i know

[taiga]

y= x^2 + 5x + 4   > derivative > dy/dx= 2x+5   when dy/dx = 0 (stationary point) >> 2x+5 = 0 >> x= -5/2
then sub that into the original equation to find the y coordinate of the stationary point

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Second one -

(2,-2) is a point hence

-2 = 4a + 2b    ....(1)

find the derivative
dy/dx = 2ax+b

when x=2 dy/dx = 3

hence

4a + b = 3  ...(2)

that's sufficient information to find the original equation (solve for a and b )

then you just use the original equation, differentiate it, make it equal to 0 and find the turning point as usual

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a hint for the third one

velocity = derivative of the displacement/time equation, the speed will just be the velocity graph with everything made positive (the modulus function of velocity - unless of course the velocity graph is always above the x axis in which case velocity and speed are the same as you are only travelling in one direction)

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