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Find the maximum or minimum of the following quadratic function: y = -10x^2 + x - 10. A. -40 B. -10 C. 10 D. -399/40

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Maximum = -399/40
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y=f(x). Take dy/dx ( f'(x) ) to get -20x+1. Critical points are where f'(x)=0. Solving: 0=-20x+1 20x=1 x=1/20 y=f(x)=f(1/20)=-9.975=-399/40
If you are in pre-calculus and have not been taught derivatives yet, put the equation in vertex form to find your min/max.

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Other answers:

Looking at f(0) to the left of critical point (=-10) and f(1) to the right (=-19), you can tell that this point is a relative maximum. Because the parabola opens down, this is the 'biggest' the function can get (absolute maximum). You could have seen all of this from graphing the function.
Thanks Guys.

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