The Student Room Group

A Level Maths Exam Question

https://www.quora.com/profile/Bravewarrior/p-163566942
Here is the question I am stuck on along with its mark scheme. I know that c=3 as that is the y-intercept but how do I get to ax^2+bx in the equation for H? And in the mark scheme they have said that 2ax+b is equal to dH/dx and I'm not quite sure what that is 😬
Well, if you were to find the maximum(/minimum, depending on sign of the x^2) of any quadratic, how would you go about that?
(edited 1 month ago)
I'm not too sure what you mean, but I assume that the equation for h being ax^2+bx+c is because h is a quadratic graph. When you differentiate a graph to get an equation (dh/dx) , you get the gradient at that point. The maximum or minimum point's gradient (dh/dx) would be zero, and they tell you that at the maximum, x would be equal to 90. So you differentiate h = ax^2 + bx + c to get dh/dx = 2ax + b. Substitute x = 90, and dh/dx = 0 into the gradient equation to get 180a + b = 0. You make it equal to zero because the gradient at that point is equal to zero. Since they also tell you when x=120, h is 27, and you understand how c is equal to 3, you can substitute it into the full equation ax^2+bx+c to get 14,400a + 120b + 3 = 27. You can then subtract that equation by three both sides to get 14,400a + 120b = 24. With this equation in terms of a and b, and the one you got by differentiating (180a + b = 0), you can solve them using simultaneous equations. If you have the Casio calculators 991 or cg50, there should be a function on your calculator that can do this for you, but it shouldn't be too hard to do it yourself. Once you find a and b, you substitute it back into ax^2 + bx + c, to get the answer the mark scheme got. Hope that helped!
Original post by tonyiptony
Well, if you were to find the maximum(/minimum, depending on sign of the x^2) of any quadratic, how would you go about that?

For a max/min, the gradient at that point would be 0, so I would differentiate the equation and set it equal to 0 πŸ™‚
Also thanks for the help!
(edited 1 month ago)
Original post by astudent6651
I'm not too sure what you mean, but I assume that the equation for h being ax^2+bx+c is because h is a quadratic graph. When you differentiate a graph to get an equation (dh/dx) , you get the gradient at that point. The maximum or minimum point's gradient (dh/dx) would be zero, and they tell you that at the maximum, x would be equal to 90. So you differentiate h = ax^2 + bx + c to get dh/dx = 2ax + b. Substitute x = 90, and dh/dx = 0 into the gradient equation to get 180a + b = 0. You make it equal to zero because the gradient at that point is equal to zero. Since they also tell you when x=120, h is 27, and you understand how c is equal to 3, you can substitute it into the full equation ax^2+bx+c to get 14,400a + 120b + 3 = 27. You can then subtract that equation by three both sides to get 14,400a + 120b = 24. With this equation in terms of a and b, and the one you got by differentiating (180a + b = 0), you can solve them using simultaneous equations. If you have the Casio calculators 991 or cg50, there should be a function on your calculator that can do this for you, but it shouldn't be too hard to do it yourself. Once you find a and b, you substitute it back into ax^2 + bx + c, to get the answer the mark scheme got. Hope that helped!

Thank youuuu so much! This is really helpful πŸ˜€πŸ˜€

Quick Reply

Latest