1. **State the problem:** We have the graph of $y=f(x)$ formed by lines AB and BC, and the graph of $y=g(x)$ formed by lines A'B' and B'C' which is obtained by applying two transformations to $y=f(x)$. We need to identify these transformations. 2. **Identify points:** - $A(1,3), B(2,5), C(4,4)$ for $y=f(x)$ - $A'(2,1), B'(4,3), C'(8,2)$ for $y=g(x)$ 3. **Find transformations from $f$ to $g$:** - Horizontal stretch: Compare $x$-coordinates: $1 \to 2$, $2 \to 4$, $4 \to 8$ means $x$ is multiplied by 2. - Vertical stretch and translation: Compare $y$-coordinates: $3 \to 1$, $5 \to 3$, $4 \to 2$ suggests $y$ is transformed by $y \to \frac{y}{2} - 0.5$. 4. **Express transformations:** - First transformation: Horizontal stretch by factor 2, $x \to 2x$. - Second transformation: Vertical stretch by factor $\frac{1}{2}$ and vertical translation down by 0.5, $y \to \frac{y}{2} - 0.5$. 5. **Transformations combined:** $g(x) = \frac{1}{2}f\left(\frac{x}{2}\right) - 0.5$. --- 6. **Transform $y=f(x)$ to $y=2f(x-1)$:** - First transformation: Horizontal translation right by 1, $x \to x-1$. - Second transformation: Vertical stretch by factor 2, $y \to 2y$. --- 7. **(a) Translate $y=x^2+2x-5$ by $\left(-\frac{1}{3},0\right)$:** - Translation left by $\frac{1}{3}$ means replace $x$ by $x+\frac{1}{3}$. - New equation: $$y = \left(x+\frac{1}{3}\right)^2 + 2\left(x+\frac{1}{3}\right) - 5$$ 8. **Expand and simplify:** $$y = x^2 + 2 \cdot x \cdot \frac{1}{3} + \left(\frac{1}{3}\right)^2 + 2x + \frac{2}{3} - 5$$ $$y = x^2 + \frac{2}{3}x + \frac{1}{9} + 2x + \frac{2}{3} - 5$$ 9. **Combine like terms:** $$y = x^2 + \left(\frac{2}{3} + 2\right)x + \left(\frac{1}{9} + \frac{2}{3} - 5\right)$$ $$y = x^2 + \frac{8}{3}x + \left(\frac{1}{9} + \frac{6}{9} - \frac{45}{9}\right)$$ $$y = x^2 + \frac{8}{3}x - \frac{38}{9}$$ --- 10. **(b) Transform $y=x^2+2x-5$ to $y=4x^2+4x-5$:** - Compare coefficients: $x^2$ term multiplied by 4, $x$ term multiplied by 2. - This corresponds to vertical stretch by factor 4 and horizontal compression by factor $\frac{1}{2}$. 11. **Single transformation:** - Horizontal compression by factor $\frac{1}{2}$: $x \to 2x$. - Vertical stretch by factor 1 (since constant term unchanged), but effectively the quadratic terms are scaled by 4. --- 12. **Find $m$ for two distinct intersections of $y=mx+1$ and $y=3x^2+2x+4$:** - Set equal: $$mx + 1 = 3x^2 + 2x + 4$$ $$3x^2 + 2x + 4 - mx - 1 = 0$$ $$3x^2 + (2 - m)x + 3 = 0$$ 13. **Discriminant for two distinct points:** $$\Delta = (2 - m)^2 - 4 \cdot 3 \cdot 3 > 0$$ $$ (2 - m)^2 > 36$$ 14. **Solve inequality:** $$|2 - m| > 6$$ 15. **Two cases:** - $2 - m > 6 \Rightarrow m < -4$ - $2 - m < -6 \Rightarrow m > 8$ **Final answer:** $$m < -4 \text{ or } m > 8$$