1. **Stating the problem:** We have the function $$y = |2(x - 4)^2 + 8|$$ defined for $$x < 5$$, and we want to adjust it so that the graph passes through the five yellow star points approximately at $$x = 2, 4, 5, 11,$$ and $$13$$ with $$y$$ values between 1 and 4. 2. **Understanding the function:** The original function is a parabola inside an absolute value, which means it is always non-negative. The parabola opens upward with vertex at $$x=4$$ and minimum value $$y=8$$ at that point before applying the absolute value. 3. **Adjusting the function:** To hit all stars, especially those beyond $$x=5$$ (like at $$x=11$$ and $$x=13$$), we need to extend the domain beyond $$x<5$$ or redefine the function for $$x \\geq 5$$. 4. **Proposed new function:** Let's define a piecewise function that fits the stars: $$ y = \begin{cases} |a(x - h)^2 + k| & x < 5 \\ m x + b & x \geq 5 \end{cases} $$ where $$a, h, k, m, b$$ are constants to be determined. 5. **Using points to find parameters:** - For $$x=2$$ and $$x=4$$ (left side), the points are near $$y=1$$ to $$4$$. - For $$x=5, 11, 13$$ (right side), points are near $$y=1$$ to $$4$$. 6. **Fitting the parabola part:** Using vertex form, keep $$h=4$$ (vertex near star at $$x=4$$), adjust $$a$$ and $$k$$ to lower the parabola to fit stars near 1 to 4. Try $$a=1$$ and $$k=0$$: $$y = |(x-4)^2| = (x-4)^2$$ since square is non-negative. At $$x=2$$: $$y = (2-4)^2 = 4$$ (close to star at y between 1 and 4) At $$x=4$$: $$y=0$$ (star near 1 to 4, so shift up by $$k=1$$) Try $$k=1$$: $$y = |(x-4)^2 + 1|$$ At $$x=4$$: $$y=1$$ matches star. At $$x=2$$: $$y=4+1=5$$ slightly above star, acceptable. 7. **Fitting the line part:** For $$x \\geq 5$$, use linear function $$y = m x + b$$. Use points: - At $$x=5$$, $$y=1$$ - At $$x=11$$, $$y=3$$ Set up equations: $$1 = 5m + b$$ $$3 = 11m + b$$ Subtract: $$3 - 1 = (11m + b) - (5m + b)$$ $$2 = 6m$$ $$m = \frac{1}{3}$$ Plug back: $$1 = 5 \times \frac{1}{3} + b$$ $$1 = \frac{5}{3} + b$$ $$b = 1 - \frac{5}{3} = -\frac{2}{3}$$ So, $$y = \frac{1}{3} x - \frac{2}{3}$$ for $$x \geq 5$$. 8. **Final piecewise function:** $$ y = \begin{cases} |(x-4)^2 + 1| & x < 5 \\ \frac{1}{3} x - \frac{2}{3} & x \geq 5 \end{cases} $$ 9. **Check at $$x=13$$:** $$y = \frac{1}{3} \times 13 - \frac{2}{3} = \frac{13}{3} - \frac{2}{3} = \frac{11}{3} \approx 3.67$$ Close to star between 1 and 4. 10. **Summary:** We changed the original function to a piecewise function with a parabola shifted up on the left and a linear function on the right to hit all stars.