1. The problem is to find the equation of a function similar to the previous one you sent. Since the previous function is not provided here, let's assume a general quadratic function for demonstration: $$y = ax^2 + bx + c$$. 2. To find the specific function, we need points or conditions. For example, if the previous function was $$y = 2x^2 - 3x + 1$$, then the similar function might have the same form but different coefficients. 3. Let's say the similar function is $$y = 3x^2 - 4x + 2$$. 4. This function is quadratic, and we can analyze its intercepts and extrema. 5. The y-intercept is found by evaluating $$y$$ at $$x=0$$: $$y = 3(0)^2 - 4(0) + 2 = 2$$. 6. The x-intercepts are found by solving $$3x^2 - 4x + 2 = 0$$ using the quadratic formula: $$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 24}}{6} = \frac{4 \pm \sqrt{-8}}{6}$$. 7. Since the discriminant is negative, there are no real x-intercepts. 8. The vertex (extremum) is at $$x = -\frac{b}{2a} = -\frac{-4}{2 \cdot 3} = \frac{4}{6} = \frac{2}{3}$$. 9. The y-value at the vertex is $$y = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) + 2 = 3 \cdot \frac{4}{9} - \frac{8}{3} + 2 = \frac{12}{9} - \frac{8}{3} + 2 = \frac{4}{3} - \frac{8}{3} + 2 = -\frac{4}{3} + 2 = \frac{2}{3}$$. 10. So the vertex is at $$\left(\frac{2}{3}, \frac{2}{3}\right)$$, which is a minimum since $$a=3 > 0$$. Final answer: The similar quadratic function is $$y = 3x^2 - 4x + 2$$ with y-intercept 2, no real x-intercepts, and a minimum vertex at $$\left(\frac{2}{3}, \frac{2}{3}\right)$$.