The question provided involves solving a system of linear equations by graphing. The two equations given are:

1. $y = x^2 - 6x + 8$ (a quadratic equation)
2. $y = -x + 4$ (a linear equation)

To solve graphically:

1. Graph the quadratic equation $y = x^2 - 6x + 8$:

   • This is a parabola. To find its vertex and intercepts:
     • The vertex is calculated using $x = -\frac{b}{2a}$, where $a = 1$, $b = -6$. So, $x = \frac{6}{2} = 3$.
     • Substituting $x = 3$ into the equation gives $y = (3)^2 - 6(3) + 8 = 9 - 18 + 8 = -1$. Thus, the vertex is $(3, -1)$.
     • Find the x-intercepts by solving $x^2 - 6x + 8 = 0$, which factors into $(x - 4)(x - 2) = 0$. So, the x-intercepts are $x = 2$ and $x = 4$.
     • The y-intercept occurs when $x = 0$, giving $y = 8$.

2. Graph the linear equation $y = -x + 4$:

   • This is a straight line with a slope of -1 and a y-intercept of 4.

Finding the intersection points:

• The intersection points of the two graphs represent the solutions to the system of equations.

• Setting $x^2 - 6x + 8 = -x + 4$, rearrange to: $x^2 - 5x + 4 = 0$ Factor this equation: $(x - 4)(x - 1) = 0$ So, $x = 4$ and $x = 1$.

• Substituting these values back into the linear equation $y = -x + 4$:

  • For $x = 4$, $y = -4 + 4 = 0$. Thus, one solution is $(4, 0)$.
  • For $x = 1$, $y = -1 + 4 = 3$. Thus, the other solution is $(1, 3)$.

Solution:

The system has two solutions: $(4, 0)$ and $(1, 3)$.

Thus, the correct answer is C. (1,3) and (4,0).

Let me know if you want more details on the solution! Here are 5 related questions to deepen your understanding:

1. How do you determine the vertex of a quadratic equation?
2. How do you find the x- and y-intercepts of a quadratic function?
3. How do you solve a system of equations algebraically?
4. What is the significance of the intersection points of two graphs?
5. How can you verify solutions graphically and algebraically?

Tip: Always double-check solutions by substituting them back into the original equations!