# Estimate the instantaneous rate of change of the function: `-2x^(2)-8` at the point (0,-8). Please use a non-derivative way of approaching this (ie. limits).

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`f(x)=-2x^2-8`

The instantaneous rate of change of the function at `(a,f(a))=(0,-8)` is given by:

`f'(a)` =`lim_(h->o)` `(f(a+h)-f(a))/h`

=`lim_(h->o)` `(f(0+h)-f(0))/h`

=`lim_(h->o)` `([-2*(0+h)^2-8]-(-8))/h`

=`lim_(h->o)` `(-2h^2-8+8)/h`

=`lim_(h->o)` `-2h`

Substituting `h=0` gives:

`f'(a)=-2*0`

`=0` `rarr` **answer**.