1. **State the problem:** We are given a parabola with vertex at $(-6, 2)$ and a point $(-8, -4)$ on the parabola. We want to find the equation of the parabola in vertex form, zero form, and standard form. 2. **Vertex form:** The vertex form of a parabola is given by: $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. Here, $h = -6$ and $k = 2$, so: $$y = a(x + 6)^2 + 2$$ 3. **Find $a$ using the point $(-8, -4)$:** Substitute $x = -8$ and $y = -4$: $$-4 = a(-8 + 6)^2 + 2$$ $$-4 = a(-2)^2 + 2$$ $$-4 = 4a + 2$$ Subtract 2 from both sides: $$-4 - 2 = 4a$$ $$-6 = 4a$$ Divide both sides by 4: $$a = \frac{\cancel{-6}}{\cancel{4}} = -\frac{3}{2}$$ 4. **Vertex form equation:** $$y = -\frac{3}{2}(x + 6)^2 + 2$$ 5. **Zero form:** The zero form is: $$y = a(x - s)(x - t)$$ We know $a = -\frac{3}{2}$ and one root is $s = -6$ (since vertex is at $x = -6$ and parabola is symmetric, the vertex is midpoint of roots). Find the other root $t$ using the point $(-8, -4)$: The vertex $x$-coordinate is midpoint: $$-6 = \frac{-8 + t}{2}$$ Multiply both sides by 2: $$-12 = -8 + t$$ Add 8 to both sides: $$t = -12 + 8 = -4$$ 6. **Zero form equation:** $$y = -\frac{3}{2}(x + 8)(x + 4)$$ 7. **Standard form:** Expand zero form: $$y = -\frac{3}{2}(x^2 + 4x + 8x + 32)$$ $$y = -\frac{3}{2}(x^2 + 12x + 32)$$ Distribute $-\frac{3}{2}$: $$y = -\frac{3}{2}x^2 - 18x - 48$$ **Final answers:** - Vertex form: $$y = -\frac{3}{2}(x + 6)^2 + 2$$ - Zero form: $$y = -\frac{3}{2}(x + 8)(x + 4)$$ - Standard form: $$y = -\frac{3}{2}x^2 - 18x - 48$$