Precalculus (6th Edition) Blitzer

a. $8$ b. $x^2-6x-19$ c. $x^2+10x-3$
a. Given $g(x)=x^2-10x-3$, we have $g(-1)=(-1)^2-10(-1)-3=1+10-3=8$ b. Similarly, $g(x+2)=(x+2)^2-10(x+2)-3=1+10-3=x^2+4x+4-10x-20-3=x^2-6x-19$ c. And $g(-x)=(-x)^2-10(-x)-3=x^2+10x-3$