1. **State the problem:** Solve the system of equations by graphing: $$y = (x - 1)^2 + 1$$ $$y = \frac{3}{5}x - 1$$ Find the points where the parabola and the line intersect. 2. **Graphing the quadratic equation:** - The quadratic is in vertex form: $$y = (x - h)^2 + k$$ where the vertex is at $$(h, k)$$. - Here, vertex is at $$(1, 1)$$. - Plot the vertex point $$(1, 1)$$. - Choose another point to the right or left, for example, at $$x=2$$: $$y = (2 - 1)^2 + 1 = 1^2 + 1 = 2$$ - Plot point $$(2, 2)$$. - By symmetry, plot $$(0, 2)$$ as well. 3. **Graphing the linear equation:** - The line is $$y = \frac{3}{5}x - 1$$. - Find intercepts: - When $$x=0$$, $$y = -1$$, so plot $$(0, -1)$$. - When $$y=0$$, solve $$0 = \frac{3}{5}x - 1$$: $$\frac{3}{5}x = 1 \Rightarrow x = \frac{5}{3} \approx 1.67$$ - Plot $$(1.67, 0)$$. - Draw the line through these points. 4. **Find intersection points (solutions):** - Set the equations equal: $$(x - 1)^2 + 1 = \frac{3}{5}x - 1$$ - Simplify: $$(x - 1)^2 + 1 + 1 = \frac{3}{5}x$$ $$(x - 1)^2 + 2 = \frac{3}{5}x$$ - Expand: $$x^2 - 2x + 1 + 2 = \frac{3}{5}x$$ $$x^2 - 2x + 3 = \frac{3}{5}x$$ - Multiply both sides by 5 to clear fraction: $$5x^2 - 10x + 15 = 3x$$ - Bring all terms to one side: $$5x^2 - 10x + 15 - 3x = 0$$ $$5x^2 - 13x + 15 = 0$$ 5. **Solve quadratic equation:** - Use quadratic formula: $$x = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 5 \cdot 15}}{2 \cdot 5} = \frac{13 \pm \sqrt{169 - 300}}{10} = \frac{13 \pm \sqrt{-131}}{10}$$ - Since discriminant is negative, no real solutions. 6. **Conclusion:** - The parabola and line do not intersect. - Therefore, there are no solutions to the system. **Answer:** There are no solutions to the system. **How to graph:** - Plot the vertex and points of the parabola. - Plot intercepts and draw the line. - Observe no intersection points.