1. **Problem statement:** We have a quadratic function $f(x) = 2x^2 - 6x + 3$ and a point $V = \left(\frac{3}{2}, -\left(\frac{3}{2}\right)^2\right)$ claimed to be the vertex of the parabola. We want to verify which statements about $V$ are true. 2. **Recall the vertex formula for a quadratic function:** For $f(x) = ax^2 + bx + c$, the vertex $V$ has coordinates $$\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right).$$ 3. **Calculate the vertex coordinates:** - Here, $a=2$, $b=-6$, $c=3$. - Compute $x$-coordinate of vertex: $$x_V = -\frac{b}{2a} = -\frac{-6}{2 \times 2} = \frac{6}{4} = \frac{3}{2} = 1.5.$$ - Compute $y$-coordinate of vertex: $$f\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^2 - 6\left(\frac{3}{2}\right) + 3 = 2 \times \frac{9}{4} - 9 + 3 = \frac{18}{4} - 6 = 4.5 - 6 = -1.5.$$ 4. **Check the given point $V$:** The user states $V = \left(\frac{3}{2}, -\left(\frac{3}{2}\right)^2\right) = \left(1.5, -2.25\right)$. However, from the function, the vertex $y$-value is $-1.5$, not $-2.25$. So the given $y$-coordinate is incorrect for the vertex of this parabola. 5. **Monotonicity change:** For a parabola opening upwards ($a=2>0$), the function decreases before the vertex and increases after it. So the vertex is the point where monotonicity changes from decreasing to increasing. This statement is true. 6. **Coordinates of $V$ as $[-b/(2a), f(-b/(2a))]$:** This is the formula for the vertex. Since the given $y$-coordinate does not match $f(-b/(2a))$, the point $V$ as given is not the vertex. But the statement about the formula itself is true. 7. **Convexity and concavity:** A parabola $f(x) = ax^2 + bx + c$ is always convex if $a>0$ or concave if $a<0$. It does not change concavity. So the statement that $V$ is the point where $f$ changes from convex to concave is false. **Final answers:** - Statement 1: True - Statement 2: True (about the formula, but the given $V$ coordinates do not match the vertex) - Statement 3: False