So first thing I would do is recognise the usefulness of the 2 points they gave you and sub them into the equation at the top, such that when x=2, y=0 and when x=0, y=-14 now from this you can set up 2 simultaneous equations and solve one for c and then sub it in to find b, from then you have the full equation and can differentiate and find whay x equals when dy/dx =0 which is the turning point.

Reply 5

Litlokhei

3

Original post by M4cc4n4

So first thing I would do is recognise the usefulness of the 2 points they gave you and sub them into the equation at the top, such that when x=2, y=0 and when x=0, y=-14 now from this you can set up 2 simultaneous equations and solve one for c and then sub it in to find b, from then you have the full equation and can differentiate and find whay x equals when dy/dx =0 which is the turning point.

but how to form a simultaneous equations?

Reply 6

hobba

9

Original post by idek23

I'm unsure what to do with this information given, could someone help me out with this please. :biggrin:

I dont know what grade you are in, but you need calculus for this one. You find the point where the derivative is zero ie 2x + b = 0 or x = -b/2. To find b and c we note that when x = 0, y = -14 so -14 = c. Similarly for b when y = 0, x = 2 so 0 = 4 + 2b -14. 0 = 2b - 10. 2b = 10 or b= 5. So the equation is y = x^2 + 5x - 14. x = -b/2 = -5/2 and y = (5/2)^2 - 5*5/2 + 5 = 25/4 - 25/2 + 5 = -25/4 + 5 = 5(-5/4 +1) = 5( -5 + 4)/4 = -5/4. So the point is (-5/2, -5/4). Unless of course I have made a mistake which these days happens more than I would like - but I am sure you see how its done - and yes I calculated the y point as well just for the heck of it.

Find the turning point of the quadratic

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