1. **State the problem:** We are given the zeros, axis of symmetry, max or min value, and vertex of a quadratic function. We need to verify these characteristics for the first quadratic. 2. **Recall the formulas and rules:** - Zeros (roots) are the x-values where the parabola crosses the x-axis. - The axis of symmetry is a vertical line through the vertex, given by $x = h$ if the vertex is $(h,k)$. - The vertex is the point $(h,k)$ where the parabola attains its maximum or minimum value. - If the parabola opens downward, the vertex is a maximum; if upward, it is a minimum. 3. **Given data for problem 1:** - Zeros: $(-4,0)$ and $(0,0)$ - Axis of symmetry: $x = 0$ - Max or Min: $y = 4$ - Vertex: $(-2,4)$ 4. **Check the axis of symmetry:** The axis of symmetry should be the vertical line through the vertex's x-coordinate. The vertex is at $(-2,4)$, so axis of symmetry should be $x = -2$, but given is $x = 0$. 5. **Check zeros and vertex:** The zeros are at $x = -4$ and $x = 0$, so the midpoint between zeros is $\frac{-4 + 0}{2} = -2$, which matches the vertex's x-coordinate. 6. **Conclusion:** - The axis of symmetry is incorrectly stated as $x=0$; it should be $x = -2$. - The vertex is at $(-2,4)$, which is a maximum since the parabola opens downward (given max $y=4$). 7. **Equation of the quadratic:** Using zeros, the factored form is: $$y = a(x + 4)(x - 0) = a(x + 4)x$$ Using vertex to find $a$: $$4 = a(-2 + 4)(-2) = a(2)(-2) = -4a \implies a = -1$$ 8. **Final equation:** $$y = -1 \cdot x(x + 4) = -x^2 - 4x$$ **Answer:** - Zeros: $-4$ and $0$ - Axis of symmetry: $x = -2$ - Max value: $y = 4$ - Vertex: $(-2,4)$ - Equation: $y = -x^2 - 4x$