1. The problem is to analyze the quadratic function given by the equation $y = kx^2 + 2x + 3$. 2. This is a quadratic function of the form $y = ax^2 + bx + c$, where $a = k$, $b = 2$, and $c = 3$. 3. Important rules for quadratic functions: - The graph is a parabola. - If $a > 0$, the parabola opens upwards; if $a < 0$, it opens downwards. - The vertex of the parabola is at $x = -\frac{b}{2a}$. - The y-intercept is at $y = c$. 4. To find the vertex: $$x = -\frac{2}{2k} = -\frac{1}{k}$$ 5. Substitute $x = -\frac{1}{k}$ back into the function to find the y-coordinate of the vertex: $$y = k\left(-\frac{1}{k}\right)^2 + 2\left(-\frac{1}{k}\right) + 3 = k\frac{1}{k^2} - \frac{2}{k} + 3 = \frac{1}{k} - \frac{2}{k} + 3 = -\frac{1}{k} + 3$$ 6. The vertex is at $$\left(-\frac{1}{k}, -\frac{1}{k} + 3\right)$$. 7. The y-intercept is when $x=0$: $$y = k(0)^2 + 2(0) + 3 = 3$$ 8. The x-intercepts (roots) can be found by solving: $$kx^2 + 2x + 3 = 0$$ Using the quadratic formula: $$x = \frac{-2 \pm \sqrt{4 - 12k}}{2k}$$ 9. The discriminant is $\Delta = 4 - 12k$. - If $\Delta > 0$, two real roots. - If $\Delta = 0$, one real root. - If $\Delta < 0$, no real roots. Summary: - Vertex: $$\left(-\frac{1}{k}, -\frac{1}{k} + 3\right)$$ - Y-intercept: 3 - X-intercepts depend on $k$ and are given by $$x = \frac{-2 \pm \sqrt{4 - 12k}}{2k}$$