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# if the equation is h= -2x^2 + 12x -10how do I find the max height?

One other way to solve this question is finding the derivative $h=-2x^2+12x-10$ $h'=-4x+12$ now we have to find when this function will be zero $-4x+12=0$ $\boxed{\boxed{x=3}}$ now we just replace this value at our initial function $h=-2x^2+12x-10$ $h_{max}=-2*(3)^2+12*3-10$ $h_{max}=-18+36-10$ $\boxed{\boxed{h_{max}=8}}$
The maximum height is the ordinate value of the vertex of the parabola, ie: yV Calculating yV: $y_V=\frac{-\Delta}{4a}\\ \\ y_V=-[\frac{12^2-4*(-2)*(-10)]}{4*(-2)}=\frac{-(144-80)}{-8}=\frac{-64}{-8}=8$