To solve the problem, we are looking for a line in the form $y = mx + c$ that allows the quadratic equation $x^2 - 7x + 4 = 0$ to be solved using the intersection points between this line and the given parabola $y = x^2 - 3x - 1$.

Steps:

1. Rewrite $x^2 - 7x + 4 = 0$ as a system: The equation $x^2 - 7x + 4 = 0$ represents a parabola equal to some other line $y$.

   Start by rearranging the quadratic equation: $x^2 - 7x + 4 = y$ To solve for intersection points, compare this to the given equation: $x^2 - 3x - 1 = y$

2. Set up equality: Since $x^2 - 3x - 1 = x^2 - 7x + 4$, cancel out $x^2$ (it appears in both): $-3x - 1 = -7x + 4$

3. Solve for $x$: Rearrange terms: $4x = 5$ $x = \frac{5}{4}$

4. Find corresponding $y$: Substitute $x = \frac{5}{4}$ into one of the equations for $y$. Using $y = x^2 - 7x + 4$: $y = \left(\frac{5}{4}\right)^2 - 7\left(\frac{5}{4}\right) + 4$ Simplify: $y = \frac{25}{16} - \frac{35}{4} + 4 = \dots\!! To correctly approach this, instead of solving intersection points directly, we determine the required **line** that represents the equation$x^2 - 7x + 4 = 0$$ as another function:

Step-by-Step Solution:

1. Rewriting $x^2 - 7x + 4 = 0$ as a system: The new parabola can be represented as: [ y = x^2 - 7x + 4$$

2. Find the line to add to the graph: Since we are finding where this new parabola intersects the existing one, $y = x^2 - 3x - 1$, we subtract the equations: $(x^2 - 7x + 4) - (x^2 - 3x - 1) = 0$

   Simplify: $-7x + 4 + 3x + 1 = 0$

   Combine like terms: $-4x + 5 = 0$

   Solve for $y$: $y = 4x - 5$

Final Answer:

The line that should be added to the graph is: $y = 4x - 5$

Would you like further clarification or a visual explanation? Here are related questions you might consider:

1. What does the graph of $y = 4x - 5$ look like relative to the parabola?
2. How do we interpret the intersection points between the parabola and this line?
3. Could we have solved this using graphing instead of algebraic manipulation?
4. How does changing the constant $c$ in $y = mx + c$ affect the solution?
5. How would this method change if the quadratic had a leading coefficient other than 1?

Tip: Always align your equations carefully when setting up intersections—they represent simultaneous solutions!