1. **State the problem:** We have a quadratic function $y = x^2 + bx + c$ with points $P(-3, -9)$ and $Q(2, -4)$ on the curve. We need to find $b$ and $c$, the y-intercept, and the x-intercepts. 2. **Form equations using the points:** Substitute $P(-3, -9)$ into $y = x^2 + bx + c$: $$-9 = (-3)^2 + b(-3) + c = 9 - 3b + c$$ Rearranged: $$-9 = 9 - 3b + c$$ $$-9 - 9 = -3b + c$$ $$-18 = -3b + c$$ Substitute $Q(2, -4)$: $$-4 = 2^2 + 2b + c = 4 + 2b + c$$ Rearranged: $$-4 - 4 = 2b + c$$ $$-8 = 2b + c$$ 3. **Write the system of equations:** $$\begin{cases} -18 = -3b + c \\ -8 = 2b + c \end{cases}$$ 4. **Solve the system:** Subtract second from first: $$-18 - (-8) = (-3b + c) - (2b + c)$$ $$-10 = -3b + c - 2b - c$$ $$-10 = -5b$$ Divide both sides by $-5$: $$\cancel{-10} / \cancel{-5} = \cancel{-5b} / \cancel{-5}$$ $$2 = b$$ Substitute $b=2$ into $-8 = 2b + c$: $$-8 = 2(2) + c$$ $$-8 = 4 + c$$ $$c = -8 - 4 = -12$$ 5. **Y-intercept:** The curve intersects the y-axis where $x=0$: $$y = 0^2 + 2(0) + (-12) = -12$$ So the y-intercept is $(0, -12)$. 6. **Find x-intercepts:** Set $y=0$: $$0 = x^2 + 2x - 12$$ Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(1)(-12)}}{2} = \frac{-2 \pm \sqrt{4 + 48}}{2} = \frac{-2 \pm \sqrt{52}}{2}$$ Simplify: $$\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}$$ So: $$x = \frac{-2 \pm 2\sqrt{13}}{2} = -1 \pm \sqrt{13}$$ Calculate approximate values: $$x_1 = -1 + 3.6 = 2.6$$ $$x_2 = -1 - 3.6 = -4.6$$ **Final answers:** - $b = 2$ - $c = -12$ - Y-intercept: $(0, -12)$ - X-intercepts: $(2.6, 0)$ and $(-4.6, 0)$ (to one decimal place)