1. Let's analyze the given graph description to understand the function $y = h(x)$. 2. The graph has two connected curved segments separated near $x = -2$: - The first segment is on the left with a concave downward shape passing through points approximately $(-7, 7)$ to an open circle near $(-2, 4)$. - The second segment is on the right with a concave upward shape starting at a filled circle near $(-2, 2)$ and passing through $(0, 5)$ and $(5, 7)$. 3. From the open and filled circle at $x=-2$, we see the function is **not continuous** at $x=-2$. 4. Approximate forms: - Left side: Since the curve is concave downward and ends lower at $x=-2$, we can model it approximately as a quadratic with negative leading coefficient. - Right side: A concave upward quadratic curve rising from $(-2,2)$ through $(0,5)$ and $(5,7)$. 5. Using points $(−2,2)$, $(0,5)$, $(5,7)$ on right segment, find quadratic $y = ax^2 + bx + c$: - At $x=0$: $c=5$ - At $x=-2$: $4a - 2b + 5 = 2$ so $4a - 2b = -3$ - At $x=5$: $25a + 5b + 5 = 7$ so $25a + 5b = 2$ 6. Solve system: - Multiply first by 5: $20a - 10b = -15$ - Multiply second by 2: $50a + 10b = 4$ - Add: $70a = -11$, thus $a = -\frac{11}{70}$ - From $4a - 2b = -3$ substitute $a$: $4(-\frac{11}{70}) - 2b = -3 \Rightarrow -\frac{44}{70} - 2b = -3$ - $-2b = -3 + \frac{44}{70} = -3 + \frac{22}{35} = -\frac{105}{35} + \frac{22}{35} = -\frac{83}{35}$ - So $b = \frac{83}{70}$ 7. Thus, right side quadratic is: $$h(x) = -\frac{11}{70}x^2 + \frac{83}{70}x + 5, \quad x \geq -2$$ 8. Left side curve is concave downward ending near $(-2,4)$ with an open circle. Passing through $(-7,7)$. Assume quadratic $y = A(x+2)^2 + B$ with vertex form for fitting: - At $x= -7$: $7 = A(-7 + 2)^2 + B = 25A + B$ - At $x= -2$: function approaches 4 but open circle means $h(-2)$ does not equal 4, so limit is 4. 9. Because of open circle at $(-2,4)$, left function does not include $x=-2$ value, so the function limit is 4 there. Approximate $B=4$. 10. Using $7=25A +4$ gives $A = \frac{3}{25}$, which is positive, so curve would be concave upward, contradicting description. Instead, try $y = D - E(x+2)^2$ (negative sign for concave downward), passing through $(-7,7)$ and approaching 4 at $x=-2$: - At $x = -2$: $y = D - E(0) = D$ approx 4, so $D=4$ - At $x = -7$: $7 = 4 - E(5)^2 = 4 - 25E$ so $-25E = 3$ and $E = -\frac{3}{25}$, negative, which contradicts the form. 11. Check form $y = Mx^2 + Nx + P$, with points $(-7,7)$ and limit near $(-2,4)$: - Let $x=-2$ limit: $M(-2)^2 + N(-2) + P = 4$ or $4M - 2N + P =4$ - At $x=-7$: $49M -7N + P=7$ 12. Let $P=4$ arbitrarily, then - $4M - 2N + 4 =4 \Rightarrow 4M - 2N =0$ - $49M -7N +4=7 \Rightarrow 49M -7N=3$ 13. Multiply first by 3.5 to eliminate $N$: - $14M -7N=0$ - Subtract from $49M -7N=3$ gives $35M=3$ so $M=\frac{3}{35}$ - From $4M - 2N=0$ gives $2N =4M = \frac{12}{35}$ so $N=\frac{6}{35}$ 14. Left function: $$h(x) = \frac{3}{35}x^2 + \frac{6}{35}x + 4, \quad x < -2$$ 15. Summary: $$h(x) = \begin{cases} \frac{3}{35}x^2 + \frac{6}{35}x + 4, & x < -2 \\ -\frac{11}{70}x^2 + \frac{83}{70}x + 5, & x \ge -2 \end{cases}$$ 16. The function is discontinuous at $x=-2$, with $h(-2) = -\frac{11}{70}(-2)^2 + \frac{83}{70}(-2) + 5 = 2$ but left side limit is 4 (open circle).