1. The problem is to create a graph of a parabola similar to the one described, which opens upwards with a vertex near (1.5, 0). 2. The general form of a parabola is given by the quadratic function $$y = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. 3. The vertex of the parabola can be found using the formula $$x = -\frac{b}{2a}$$. Given the vertex is at $x = 1.5$, we can use this to find $b$ if $a$ is known. 4. To have the vertex at $(1.5, 0)$, the function must satisfy $$y(1.5) = 0$$. 5. For example, choose $a = 1$ (which makes the parabola open upwards). Then, $$1.5 = -\frac{b}{2 \times 1} \Rightarrow b = -3$$. 6. Now, use the vertex point to find $c$: $$0 = 1 \times (1.5)^2 - 3 \times 1.5 + c \Rightarrow 0 = 2.25 - 4.5 + c \Rightarrow c = 2.25$$. 7. The quadratic function is $$y = x^2 - 3x + 2.25$$. 8. This function will produce a parabola opening upwards with vertex at $(1.5, 0)$, spanning the x and y ranges described. 9. To graph this, plot points for values of $x$ from 0 to 12 and calculate corresponding $y$ values using the function. 10. Use graphing tools or software like Desmos to visualize the curve.