Example. $\mathrm{y}=-2 \mathrm{x}^{2}+3 \mathrm{x}+4$.
Solving $\mathrm{y}=4$ gives the required straight line. Thus, $-2 x^{2}+3 x+4=4$, solving for $x$ yields $x=0$, $\mathrm{x}=\frac{3}{2}$, which are the x-coordinates of the two intersection points of the parabola and the line $\mathrm{y}=4$. The equation of the axis of symmetry is $\mathrm{x}=\frac{3}{4}$, and the vertex is $\left(\frac{3}{4}, \frac{41}{8}\right)$.

Since the parabola passes through the points $(0,4)$ and $\left(\frac{3}{2}, 4\right)$, a rough sketch can be drawn, as shown in Figure 2.

From this example, it can be seen that when $a$, $b$, and $c$ are rational numbers, using this method to find the vertex of the parabola and draw a rough sketch is very convenient.

Finally, for the special form of the quadratic function $\mathrm{y}=\mathrm{ax} \mathrm{x}^{2}$, the parabola intersects the line $\mathrm{y}=0$ at only one point $(0,0)$, at which the equation of the axis of symmetry is $\mathbf{x}=0$, and the vertex is $(0,0)$.

To draw the graph, take two more points on the $x$-axis, such as $x= \pm 1$, giving $(-1, a)$ and $(1, a)$. This allows the rough sketch of the parabola to be made.

(Translated from the American "Mathematics Teacher", March 1979 issue)