1. **State the problem:** We are given the function $y = x(x-4) - 5$ and asked to draw its graph for $-3 \leq x \leq 4$. Then, we need to use this graph to answer two quadratic equations: (a) $x^2 - 5x - 3 = 0$ (b) $x^2 - 3x + 4 = 0$ 2. **Rewrite the function:** Expand $y = x(x-4) - 5$: $$y = x^2 - 4x - 5$$ 3. **Graph features:** This is a quadratic function with: - $a = 1$, $b = -4$, $c = -5$ - Vertex at $x = -\frac{b}{2a} = -\frac{-4}{2 \times 1} = 2$ - Vertex $y$ value: $y = 2^2 - 4 \times 2 - 5 = 4 - 8 - 5 = -9$ - The parabola opens upwards since $a > 0$. 4. **Use the graph to solve the equations:** The graph of $y = x^2 - 4x - 5$ is close to the equations given but not exactly the same. However, we can compare the roots of the given equations to the graph of $y = x^2 - 4x - 5$ to understand their solutions. 5. **Solve (a) $x^2 - 5x - 3 = 0$ algebraically:** Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{25 + 12}}{2} = \frac{5 \pm \sqrt{37}}{2}$$ Approximate roots: $$x \approx \frac{5 \pm 6.08}{2}$$ So, $$x_1 \approx \frac{5 + 6.08}{2} = 5.54$$ $$x_2 \approx \frac{5 - 6.08}{2} = -0.54$$ 6. **Solve (b) $x^2 - 3x + 4 = 0$ algebraically:** Discriminant: $$\Delta = (-3)^2 - 4 \times 1 \times 4 = 9 - 16 = -7 < 0$$ Since discriminant is negative, no real roots. 7. **Summary:** - Equation (a) has two real roots approximately $-0.54$ and $5.54$. - Equation (b) has no real roots. Since the graph of $y = x^2 - 4x - 5$ is similar but shifted compared to the equations, the roots differ slightly.