1. **State the problem:** Graph the quadratic function $y = 6x^2 + 5x - 4$ and find the x-intercepts, y-intercept, axis of symmetry, and vertex. 2. **Formula and rules:** - The quadratic function is in standard form $y = ax^2 + bx + c$ where $a=6$, $b=5$, and $c=-4$. - The x-intercepts are found by solving $6x^2 + 5x - 4 = 0$. - The y-intercept is the value of $y$ when $x=0$, which is $c$. - The axis of symmetry is given by $x = -\frac{b}{2a}$. - The vertex is at $(x, y)$ where $x$ is the axis of symmetry and $y$ is $f(x)$. 3. **Find the x-intercepts:** Solve $6x^2 + 5x - 4 = 0$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Substitute values: $$x = \frac{-5 \pm \sqrt{5^2 - 4 \times 6 \times (-4)}}{2 \times 6} = \frac{-5 \pm \sqrt{25 + 96}}{12} = \frac{-5 \pm \sqrt{121}}{12}$$ $$x = \frac{-5 \pm 11}{12}$$ Two solutions: $$x_1 = \frac{-5 + 11}{12} = \frac{6}{12} = \frac{1}{2}$$ $$x_2 = \frac{-5 - 11}{12} = \frac{-16}{12} = -\frac{4}{3}$$ 4. **Find the y-intercept:** At $x=0$, $y = 6(0)^2 + 5(0) - 4 = -4$. 5. **Find the axis of symmetry:** $$x = -\frac{b}{2a} = -\frac{5}{2 \times 6} = -\frac{5}{12}$$ 6. **Find the vertex:** Calculate $y$ at $x = -\frac{5}{12}$: $$y = 6\left(-\frac{5}{12}\right)^2 + 5\left(-\frac{5}{12}\right) - 4 = 6 \times \frac{25}{144} - \frac{25}{12} - 4 = \frac{150}{144} - \frac{25}{12} - 4$$ Simplify fractions: $$\frac{150}{144} = \frac{25}{24}, \quad \frac{25}{12} = \frac{50}{24}, \quad 4 = \frac{96}{24}$$ So: $$y = \frac{25}{24} - \frac{50}{24} - \frac{96}{24} = \frac{25 - 50 - 96}{24} = \frac{-121}{24}$$ 7. **Summary:** - x-intercepts: $\left(\frac{1}{2}, 0\right)$ and $\left(-\frac{4}{3}, 0\right)$ - y-intercept: $(0, -4)$ - Axis of symmetry: $x = -\frac{5}{12}$ - Vertex: $\left(-\frac{5}{12}, -\frac{121}{24}\right)$ This completes the analysis and graphing information for the quadratic function.