if the equation is h= -2x^2 + 12x -10

how 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:

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

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