1. **State the problem:** We need to find the quadratic equation $y = ax^2 + bx + c$ given that the graph has a turning point at $(-2,4)$ and passes through the point $(0,8)$. 2. **Use the vertex form of a quadratic:** The turning point (vertex) form is $$y = a(x - h)^2 + k$$ where $(h,k)$ is the vertex. Here, $h = -2$ and $k = 4$, so $$y = a(x + 2)^2 + 4$$ 3. **Use the point $(0,8)$ to find $a$:** Substitute $x=0$ and $y=8$ into the vertex form: $$8 = a(0 + 2)^2 + 4$$ $$8 = a(2)^2 + 4$$ $$8 = 4a + 4$$ 4. **Solve for $a$:** $$4a = 8 - 4$$ $$4a = 4$$ $$a = 1$$ 5. **Write the equation in vertex form:** $$y = 1(x + 2)^2 + 4 = (x + 2)^2 + 4$$ 6. **Expand to standard form $y = ax^2 + bx + c$:** $$y = (x + 2)^2 + 4 = x^2 + 4x + 4 + 4 = x^2 + 4x + 8$$ **Final answer:** $$y = x^2 + 4x + 8$$