1. The problem asks for the line in the form $y = mx + c$ that should be drawn on the graph to solve the quadratic equation $x^2 - 7x + 4 = 0$. 2. To solve $x^2 - 7x + 4 = 0$ graphically, we can rewrite it as $y = x^2 - 7x + 4$ and find where this parabola intersects the x-axis (i.e., where $y=0$). 3. Another way is to consider the original parabola $y = x^2 - 3x - 1$ and find a line $y = mx + c$ such that the intersection points of this line and the parabola correspond to the solutions of $x^2 - 7x + 4 = 0$. 4. Set the parabola equal to the line: $$x^2 - 3x - 1 = mx + c$$ Rearranged: $$x^2 - 3x - 1 - mx - c = 0$$ $$x^2 - (3 + m)x - (1 + c) = 0$$ 5. For this quadratic to be equivalent to $x^2 - 7x + 4 = 0$, the coefficients must match: - Coefficient of $x$: $-(3 + m) = -7$ so $3 + m = 7$ which gives $m = 4$ - Constant term: $-(1 + c) = 4$ so $1 + c = -4$ which gives $c = -5$ 6. Therefore, the line is: $$y = 4x - 5$$ 7. This line intersects the parabola $y = x^2 - 3x - 1$ at points where $x^2 - 7x + 4 = 0$, solving the equation graphically. Final answer: $$y = 4x - 5$$