Question

How do you evaluate the function `h(x)= .2x^2 - .6x - .25` for `h(-1/2)`?

Answer 1

`h(-1/2) = 0.1`

Explanation:

Substituting `(-1/2)` for `x` in `h(x)=0.2x^2-0.6x-0.25`
gives
`color(white)("XXX")h(-1/2) = 0.2xx(-1/2)^2-0.6xx(-1/2)-0.25`

`color(white)("XXXXXXXX")= (0.2xx1/4) - (-0.3) -0.25`

`color(white)("XXXXXXXX")=0.05+0.3-0.25`

`color(white)("XXXXXXXX")=0.1`