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