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Parts A and B
\nHere, the question takes you throught the stages needed to find the solution. The reason we differentiate is that the derivative of a function tells us its gradient at a given point, and we want to find where the function has gradient zero because when the gradient is zero we either have a maximum or a minimum point.
\nPart C
\nThe first part of this question is similar to parts A and B. The tricky bit is the second part! You need to work out the value of $t$ that produces the maximum piont but that is not the final answer - you need to use that value of $t$ to find the maximum height, which you do by substituting $t$ into the original function to find $y$.
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\n\\[ y = \\simplify{ {a}*x^2 + {b}x + {c}} \\]
\nFirstly, differentiate.
\n$\\displaystyle \\frac{dy}{dx}=$ [[1]]
\nGradient at $x=\\var{d}\\;$ is [[0]]
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\nFirstly, find the first and second derivatives $y$.
\n$\\displaystyle \\frac{dy}{dx}=$ [[2]]
\n$\\displaystyle \\frac{d^2y}{dx^2}=$ [[3]]
\n\nSecondly, find $x$ such that $\\displaystyle \\frac{dy}{dx}=0$.
\n$x$-coordinate of the turning point $=$ [[0]]
\n$y$-coordinate of the turning point $=$ [[1]]
\nThe turning point is a [[4]]
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\nAt a time $t$ seconds after the instant of projection, its height, $y$ metres, above the ground is given by the formula
\n\\[ y=\\var{z}t-\\var{w}t^2. \\]
\nCalculate the maximum height reached by the missile.
\nFirstly, differentiate.
\n$\\displaystyle \\frac{dy}{dt}=$ [[0]]
\nNow use this result and your knowledge of differentiation to find the maximum height of the missile, rounding your answer to $2$ decimal places.
\n$y=$ [[1]]
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