Verified Answer :

D3xt3R

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

\(h=-2x^2+12x-10\)

\(h_{max}=-2*(3)^2+12*3-10\)

\(h_{max}=-18+36-10\)

\(\boxed{\boxed{h_{max}=8}}\)

Ryan2

The maximum height is the ordinate value of the vertex of the parabola, ie: yV

Calculating yV:

\(y_V=\frac{-\Delta}{4a}\\ \\ y_V=-[\frac{12^2-4*(-2)*(-10)]}{4*(-2)}=\frac{-(144-80)}{-8}=\frac{-64}{-8}=8\)

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