1. We are given the quadratic function $f(x) = 2x^2 + 6x - 5$ and asked to analyze it. 2. The general form of a quadratic function is $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants. 3. Important rules: - The parabola opens upwards if $a > 0$ and downwards if $a < 0$. - The vertex of the parabola is at $x = -\frac{b}{2a}$. - The y-coordinate of the vertex is $f\left(-\frac{b}{2a}\right)$. - The axis of symmetry is the vertical line $x = -\frac{b}{2a}$. 4. Calculate the vertex: - $a = 2$, $b = 6$, $c = -5$ - $x$-coordinate of vertex: $x_v = -\frac{6}{2 \times 2} = -\frac{6}{4} = -1.5$ 5. Calculate the y-coordinate of the vertex: - $f(-1.5) = 2(-1.5)^2 + 6(-1.5) - 5$ - $= 2(2.25) - 9 - 5$ - $= 4.5 - 9 - 5 = -9.5$ 6. The vertex is at $(-1.5, -9.5)$. 7. Since $a = 2 > 0$, the parabola opens upwards. 8. Find the y-intercept by evaluating $f(0)$: - $f(0) = 2(0)^2 + 6(0) - 5 = -5$ 9. Find the x-intercepts by solving $2x^2 + 6x - 5 = 0$ using the quadratic formula: - $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4 \times 2 \times (-5)}}{2 \times 2}$ - $= \frac{-6 \pm \sqrt{36 + 40}}{4} = \frac{-6 \pm \sqrt{76}}{4}$ - $= \frac{-6 \pm 2\sqrt{19}}{4} = \frac{-3 \pm \sqrt{19}}{2}$ 10. The x-intercepts are $x = \frac{-3 + \sqrt{19}}{2}$ and $x = \frac{-3 - \sqrt{19}}{2}$. Final answer: - Vertex: $(-1.5, -9.5)$ - Opens upward - Y-intercept: $(0, -5)$ - X-intercepts: $\left(\frac{-3 + \sqrt{19}}{2}, 0\right)$ and $\left(\frac{-3 - \sqrt{19}}{2}, 0\right)$