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    Equation of graph is - y=x^3+x^2-5x-2
    1. Show that there is a turning point when x=1 and find the x co-ordinate of the turning point
    2.Find the equation of the tangent at the point (0,-2)
    3.Find the equation of the normal at the point (2,0)
    Thanks
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    (Original post by Sandrathomas)
    Equation of graph is - y=x^3+x^2-5x-2
    1. Show that there is a turning point when x=1 and find the x co-ordinate of the turning point
    2.Find the equation of the tangent at the point (0,-2)
    3.Find the equation of the normal at the point (2,0)
    Thanks
    1. Differentiate the function to get 3x^2 + 2x -5, then equate it to zero and solve teh quadratic , which should give you 2 turning points , one being x =1.

    2. sub in x= 0 into 3x^2 +2x- 5 , you get -5 , so the gradient of the tangent is -5, therefore y= -5x + c, sub in your coordiante (0, -2) to find C and that is your equation.

    3. the gradient of the normal is the negative reciprocal of the gradient of the tangent , which is 1/5 so y= 1/5x+c , sub in (2,0) to find your equation.
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    (Original post by ihatehannah)
    1. Differentiate the function to get 3x^2 + 2x -5, then equate it to zero and solve teh quadratic , which should give you 2 turning points , one being x =1.

    2. sub in x= 0 into 3x^2 +2x- 5 , you get -5 , so the gradient of the tangent is -5, therefore y= -5x + c, sub in your coordiante (0, -2) to find C and that is your equation.

    3. the gradient of the normal is the negative reciprocal of the gradient of the tangent , which is 1/5 so y= 1/5x+c , sub in (2,0) to find your equation.
    Thankyou so much, this really helped x
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    (Original post by Sandrathomas)
    Thankyou so much, this really helped x

    Heya, can I just check something?
    1. x is 1 or -5/3, so is is -5/3 the answer?
    2. y=-5x-2?
    3. y=1/5x+-2/5?
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    (Original post by Sandrathomas)
    Heya, can I just check something?
    1. x is 1 or -5/3, so is is -5/3 the answer?
    2. y=-5x-2?
    3. y=1/5x+-2/5?
    yeah thats right
 
 
 
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