If y = -x^2 + 3, what is the largest possible value of y?
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Originally Posted by magentarita If y = -x^2 + 3, what is the largest possible value of y? Again... differentiate wrt y, set equal to zero, solve for x, and put x value back into original equation! $\displaystyle \frac{dy}{dx} = -2x = 0$
Originally Posted by magentarita If y = -x^2 + 3, what is the largest possible value of y? Hello Magentarita, You are looking for the max value of the parabola with vertex (0, 3). The largest value of y is the y-coordinate of the vertex.
Originally Posted by Mush Again... differentiate wrt y, set equal to zero, solve for x, and put x value back into original equation! $\displaystyle \frac{dy}{dx} = -2x = 0$ Why did you reply using calculus? This is an algebra 2 question.
Originally Posted by masters Hello Magentarita, You are looking for the max value of the parabola with vertex (0, 3). The largest value of y is the y-coordinate of the vertex. So, basically we set x = 0 and simplify, right?
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