1. **State the problem:** We need to graph the quadratic function $$f(x) = 5x^2 + 10x - 5$$ and plot its vertex and another point on the parabola. 2. **Recall the vertex formula:** For a quadratic function $$f(x) = ax^2 + bx + c$$, the vertex $$x$$-coordinate is given by $$x = -\frac{b}{2a}$$. 3. **Calculate the vertex:** Here, $$a = 5$$ and $$b = 10$$. $$x = -\frac{10}{2 \times 5} = -\frac{10}{10} = -1$$ 4. **Find the vertex's $$y$$-coordinate:** Substitute $$x = -1$$ into the function: $$f(-1) = 5(-1)^2 + 10(-1) - 5 = 5(1) - 10 - 5 = 5 - 10 - 5 = -10$$ So, the vertex is at $$(-1, -10)$$. 5. **Plot another point:** Choose $$x = 0$$ for simplicity. $$f(0) = 5(0)^2 + 10(0) - 5 = -5$$ So, another point on the parabola is $$(0, -5)$$. 6. **Summary:** The parabola opens upwards (since $$a = 5 > 0$$), vertex at $$(-1, -10)$$, and passes through $$(0, -5)$$. This information allows you to sketch the parabola accurately on the Cartesian plane.