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Express y = 14 - 3x - x^2 in the form y = p(x + q)^2. Hence, find the maximum value of y and the value of x at which the maximum value occurs.
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\[\frac{65}{4}-\left(x+\frac{3}{2}\right)^2\]
Maximum at vertex (-3/2, 65/4) Because the coefficient of x^2 is negative, its a maximum
Thanks again! Any idea how to sketch the graph for y = 14 - 3x - x^2?
THANK YOU, Sam! (:
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