1. The problem involves a piecewise function: $$f(x) = \begin{cases} x^2 & \text{if } x > 10 \\ 3x + 1 & \text{if } x \leq 10 \end{cases}$$ 2. We are given points and sets: - Points: P(-3,6), Q(-2,3), R(0,-1), S(1,4), T(5,15), U(7,7), V(9,11) - Sets: a = \{-3, -2, -2, 0, 1, 5, 5, 7, 9, 11, 12\}, X = \{-3, -2, 0, 1, 5, 7, 9\}, Y = \{6, 3, -1, 4, 15, 7, 11\} 3. We want to verify if the points lie on the function $f(x)$ by evaluating $f(x)$ at each $x$-coordinate and comparing to the $y$-coordinate. 4. Evaluate $f(x)$ for each $x$ in the points: - For $x = -3$ (since $-3 \leq 10$), use $f(x) = 3x + 1$: $$f(-3) = 3(-3) + 1 = -9 + 1 = -8$$ Given $y=6$, point P(-3,6) does not lie on $f(x)$. - For $x = -2$ (\leq 10): $$f(-2) = 3(-2) + 1 = -6 + 1 = -5$$ Given $y=3$, point Q(-2,3) does not lie on $f(x)$. - For $x = 0$ (\leq 10): $$f(0) = 3(0) + 1 = 1$$ Given $y=-1$, point R(0,-1) does not lie on $f(x)$. - For $x = 1$ (\leq 10): $$f(1) = 3(1) + 1 = 4$$ Given $y=4$, point S(1,4) lies on $f(x)$. - For $x = 5$ (\leq 10): $$f(5) = 3(5) + 1 = 15 + 1 = 16$$ Given $y=15$, point T(5,15) does not lie on $f(x)$. - For $x = 7$ (\leq 10): $$f(7) = 3(7) + 1 = 21 + 1 = 22$$ Given $y=7$, point U(7,7) does not lie on $f(x)$. - For $x = 9$ (\leq 10): $$f(9) = 3(9) + 1 = 27 + 1 = 28$$ Given $y=11$, point V(9,11) does not lie on $f(x)$. 5. For $x > 10$, use $f(x) = x^2$: - For $x=11$: $$f(11) = 11^2 = 121$$ - For $x=12$: $$f(12) = 12^2 = 144$$ 6. Summary: Only point S(1,4) lies on the function $f(x)$ as given. 7. The function $f(x)$ is piecewise linear and quadratic, with a breakpoint at $x=10$. 8. The graph of $f(x)$ can be described as: $$y = \begin{cases} x^2 & x > 10 \\ 3x + 1 & x \leq 10 \end{cases}$$ This completes the verification of points against the function.