Given the functions :
[tex]\begin{gathered} f(x)=-2x^2-4 \\ h(x)=x^2-3x+5 \end{gathered}[/tex]
a. find f ( -1/4)
so, substitute with x = -1/4 at the function f
[tex]f(-\frac{1}{4})=-2\cdot(-\frac{1}{4})^2-4=-2\cdot\frac{1}{16}-4=-\frac{1}{8}-4=-4.125[/tex]
b. find f(-3x+2)
so, substitute with x = -3x + 2
[tex]\begin{gathered} f(-3x+2)=-2\cdot(-3x+2)^2-4 \\ =-2\cdot(9x^2-12x+4)-4 \\ =-18x^2+24x-8+4 \\ =-18x^2+24x-4 \end{gathered}[/tex]
C. find h (5x) - 4
So, the answer will be :
[tex]\begin{gathered} h(5x)-4=(5x)^2-3\cdot(5x)+5-4 \\ \end{gathered}[/tex]
d. -4h(3+k)
so, the answer will be :
[tex]-4h(3+k)=-4\cdot\lbrack(3+k)^2-3\cdot(3+k)+5\rbrack[/tex]
e. f(-x^3)
So ,the answer will be :
[tex]\begin{gathered} f(-x)^3=-2\cdot(-x^3)^2-4 \\ \end{gathered}[/tex]
f. h(-2/3)
So, the answer will be :
[tex]h(-\frac{2}{3})=(-\frac{2}{3})^2-3\cdot(-\frac{2}{3})+5=\frac{67}{9}[/tex]
