## Task B-2.7.

The functions $f(x)=-\frac{1}{4} x^{2}+3 x-2$ and $g(x)=|2 x-4|-6$ are given. What is the area of the triangle whose two vertices are the points of intersection of the graphs of functions $f$ and $g$, and the third vertex of the triangle is the point of minimum of function $g$?
## First Solution.

Let's draw the graph of the function $f(x)=-\frac{1}{4} x^{2}+3 x-2$.

First, let's determine the coordinates of its vertex.

$x_{T}=-\frac{b}{2 a}=6, \quad y_{T}=\frac{4 a c-b^{2}}{4 a}=7$

The vertex of the function $f$ is the point $T(6,7)$.

Next, let's determine a few points that belong to the graph of the function $f$ and sketch the desired graph.

| $x$ | 0 | 2 | 4 | 6 | 8 | 10 | 12 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $f(x)$ | -2 | 3 | 6 | 7 | 6 | 3 | -2 |

In the same coordinate system, let's also draw the graph of the function $g(x)=|2 x-4|-6$.

The table below shows the calculated values of the function $g$ for some arguments $x$.

| $x$ | -4 | -2 | 0 | 2 | 4 | 6 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $f(x)$ | 6 | 2 | -2 | -6 | -2 | 2 | 6 |

![](https://cdn.mathpix.com/cropped/2024_05_30_4b7629e40479ae2c2357g-16.jpg?height=659&width=788&top_left_y=1047&top_left_x=180)

The minimum point of the function $g$ is $(2,-6)$.

The intersection points of the graphs of the functions $f$ and $g$ are points that belong to both graphs, and these are the points with coordinates $(0,-2)$ and $(8,6)$.

Let's sketch a triangle with vertices $(2,-6),(0,-2)$, and $(8,6)$ and determine its area.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b7629e40479ae2c2357g-16.jpg?height=599&width=574&top_left_y=1965&top_left_x=181)

The area of the triangle is:

$P=12 \cdot 8-\frac{8 \cdot 8}{2}-\frac{12 \cdot 6}{2}-\frac{4 \cdot 2}{2}=24$ square units.

Note: A student who reads the coordinates of the intersection points from the graph without verifying that the points actually belong to both graphs can receive a maximum of 9 points.