To solve the given system of equations by graphing:

1. Equations:

   • $y = -x^2 + 2x + 4$ (a parabola)
   • $y = 2x + 4$ (a straight line)

2. Steps to solve:

   • Graph $y = -x^2 + 2x + 4$:

     • This is a downward-opening parabola because the coefficient of $x^2$ is negative.
     • Vertex: The vertex is at $x = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1$. Substituting $x = 1$ into the equation gives $y = -1^2 + 2(1) + 4 = 5$. So, the vertex is $(1, 5)$.
     • Additional points can be found by substituting other $x$-values.

   • Graph $y = 2x + 4$:

     • This is a straight line with slope $2$ and $y$-intercept $4$.
     • The points $(0, 4)$ and $(1, 6)$ can be plotted.

3. Find intersections:

   • Set the two equations equal to each other: $-x^2 + 2x + 4 = 2x + 4$ Simplify: $-x^2 = 0$ $x = 0$ Substitute $x = 0$ back into one of the equations, e.g., $y = 2x + 4$: $y = 4$ The intersection is at $(0, 4)$.

The system has one solution, $(0, 4)$.

Would you like me to verify or expand any steps?

Related Questions:

1. What are the key features of parabolas, and how do they affect graphing?
2. How do we determine the vertex and axis of symmetry of a quadratic equation?
3. What are alternative methods for solving systems of equations apart from graphing?
4. How can the slope-intercept form of a line be derived and used effectively?
5. What conditions determine the number of solutions for a system of linear and quadratic equations?

Tip: For cleaner graphing, always use precise values for the vertex and intercepts to plot curves or lines.