Question

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

how do I find the max height?

Answer 1

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

Calculating yV:


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

Answer 2

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}}`