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Parts A and B

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Here, 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.

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Part C

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The 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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Find the gradient of the curve $y$ at the point $x=\\var{d}$, giving your answer to $2$ decimal places if necessary.

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\$y = \\simplify{ {a}*x^2 + {b}x + {c}} \$

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Firstly, differentiate.

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$\\displaystyle \\frac{dy}{dx}=$ [[1]]

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Gradient at $x=\\var{d}\\;$ is [[0]]

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Find the coordinates of the turning point of the function below and state whether it is a maximum or a minimum point. Give your answers to $2$ decimal places where necessary.

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$y=\\simplify {{f}x^2+{g}x+{h}}$

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Firstly, find the first and second derivatives $y$.

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$\\displaystyle \\frac{dy}{dx}=$ [[2]]

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$\\displaystyle \\frac{d^2y}{dx^2}=$ [[3]]

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Secondly, find $x$ such that $\\displaystyle \\frac{dy}{dx}=0$.

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$x$-coordinate of the turning point $=$ [[0]]

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$y$-coordinate of the turning point $=$ [[1]]

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The turning point is a [[4]]

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An unpowered missile is launched vertically from the ground.

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At a time $t$ seconds after the instant of projection, its height, $y$ metres, above the ground is given by the formula

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\$y=\\var{z}t-\\var{w}t^2. \$

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Calculate the maximum height reached by the missile.

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Firstly, differentiate.

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$\\displaystyle \\frac{dy}{dt}=$ [[0]]

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Now use this result and your knowledge of differentiation to find the maximum height of the missile, rounding your answer to $2$ decimal places.

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$y=$ [[1]]

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Is the stationary point a maximum?

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