1. **Problem 8:** Describe the transformations for each graph from $y=f(x)$. - a) $y=f(x+4)$ shifts the graph of $f(x)$ 4 units to the left. - b) $y=f(x-4)$ shifts the graph of $f(x)$ 4 units to the right. - c) $y=f(x+4)-3$ shifts the graph 4 units left and 3 units down. - d) $y=f(x+4)+3$ shifts the graph 4 units left and 3 units up. - e) $y=f(-x)+3$ reflects the graph about the y-axis and shifts it 3 units up. - f) $y=f(-x+3)$ reflects about y-axis and shifts 3 units right (since $-x+3=-(x-3)$). - g) $y=f(4x)$ compresses the graph horizontally by a factor of 4. - h) $y=4f(x)$ stretches the graph vertically by a factor of 4. 2. **Problem 9:** Given $f(x)=x^2+4$ and $g(x)=\frac{6x}{x^2-9}$, find $(f\circ g)(4)$ and $(g\circ f)(4)$. - Compute $g(4)=\frac{6\times4}{16-9}=\frac{24}{7}$. - Then $(f\circ g)(4)=f(g(4))=f\left(\frac{24}{7}\right)=\left(\frac{24}{7}\right)^2+4=\frac{576}{49}+4=\frac{576}{49}+\frac{196}{49}=\frac{772}{49}$. - Compute $f(4)=4^2+4=16+4=20$. - Then $(g\circ f)(4)=g(f(4))=g(20)=\frac{6\times20}{400-9}=\frac{120}{391}$. 3. **Problem 10:** Given $f(x)=x^2$ and $g(x)=2x+1$, find: - a) $(f\pm g)(x)=f(x)\pm g(x)=x^2 \pm (2x+1)$. - b) $(f/g)(x)=\frac{f(x)}{g(x)}=\frac{x^2}{2x+1}$. - c) $(g\circ f)(2)=g(f(2))=g(2^2)=g(4)=2\times4+1=9$. - d) $(f\circ g)(2)=f(g(2))=f(2\times2+1)=f(5)=5^2=25$. - e) $(fg)(2)=f(2)\times g(2)=2^2 \times (2\times2+1)=4 \times 5=20$. 4. **Problem 11:** Find linear function $f(x)$ given $f(-2)=6$ and $f(1)=3$. - Slope $m=\frac{3-6}{1-(-2)}=\frac{-3}{3}=-1$. - Use point-slope form: $y-6=-1(x+2)$. - Simplify: $y=-x-2+6=-x+4$. - So $f(x)=-x+4$. 5. **Problem 12:** Given $Q=3.34b^3+1.870b^2x$ with $b$ constant. - Since $Q$ depends on $x$ linearly (term $1.870b^2x$) plus a constant term $3.34b^3$, the graph of $Q$ vs $x$ is a straight line. - Slope when $b=1$ is coefficient of $x$: $1.870 \times 1^2=1.870$. 6. **Problem 13:** Voltage $V$ and current $I$ linear relation. - Given points: $(I,V)=(4,2)$ and $(12,6)$. - Slope $m=\frac{6-2}{12-4}=\frac{4}{8}=0.5$. - Equation: $V=0.5I + c$. - Use point $(4,2)$: $2=0.5\times4 + c \Rightarrow c=0$. - So $V=0.5I$. - For $I=10$, $V=0.5\times10=5$. 7. **Problem 14:** Radioactive decay $N=100e^{-0.062t}$. - Initial amount at $t=0$ is $N=100e^0=100$ mg. - After 10 days: $N=100e^{-0.062\times10}=100e^{-0.62}$. - Approximate $e^{-0.62}\approx0.538$. - So $N\approx100\times0.538=53.8$ mg. 8. **Problem 15:** Given $1.5M=\log\left(\frac{E}{2.5\times10^{11}}\right)$, solve for $E$. - Multiply both sides by $\frac{1}{1.5}$: $M=\frac{1}{1.5}\log\left(\frac{E}{2.5\times10^{11}}\right)$. - Rewrite: $1.5M=\log\left(\frac{E}{2.5\times10^{11}}\right)$. - Exponentiate base 10: $10^{1.5M}=\frac{E}{2.5\times10^{11}}$. - Multiply both sides: $E=2.5\times10^{11} \times 10^{1.5M}=2.5\times10^{11+1.5M}$. **Final answers:** - 8: See transformations above. - 9: $(f\circ g)(4)=\frac{772}{49}$, $(g\circ f)(4)=\frac{120}{391}$. - 10a: $(f\pm g)(x)=x^2 \pm (2x+1)$. - 10b: $(f/g)(x)=\frac{x^2}{2x+1}$. - 10c: 9. - 10d: 25. - 10e: 20. - 11: $f(x)=-x+4$. - 12: Graph is a straight line; slope at $b=1$ is 1.870. - 13: $V=0.5I$, voltage at $I=10$ is 5. - 14: Initial amount 100 mg; after 10 days approx 53.8 mg. - 15: $E=2.5\times10^{11+1.5M}$.