8.2. The part of the graph of a linear function located in the second coordinate quadrant, together with the coordinate axes, forms a triangle. By what factor will its area change if the slope of the function is doubled and the y-intercept is halved?
Answer: It will decrease by eight times.

Solution. Let the original linear function be defined by the equation $y=k x+b$. From the condition of the problem, it follows that $k>0$ and $b>0$ (see Fig. 8.2). The points of intersection of its graph with the axes are: $A(0 ; b)$ and $B\left(-\frac{b}{k} ; 0\right)$. Since the triangle $A O B$ cut off by it is a right triangle, its area is $\frac{1}{2} O A \cdot O B=\frac{1}{2} b \cdot \frac{b}{k}=\frac{b^{2}}{2 k}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1c054e213dcb73e48b02g-1.jpg?height=460&width=606&top_left_y=1069&top_left_x=1316)

After the specified change in coefficients, the function becomes $y=2 k x+0.5 b$. Its graph intersects the coordinate axes at points $\mathrm{C}\left(0 ; \frac{b}{2}\right)$ and $D\left(-\frac{b}{4 k} ; 0\right)$. The area of the right triangle COD is $\frac{1}{2} O C \cdot O D=\frac{1}{2} \cdot \frac{b}{2} \cdot \frac{b}{4 k}=$ $\frac{b^{2}}{16 k}$. Therefore, the area will decrease by $\frac{b^{2}}{2 k}: \frac{b^{2}}{16 k}=8$ times.

Grading Criteria. «+» A complete and well-reasoned solution is provided «士» A generally correct reasoning is provided, containing minor gaps or inaccuracies that did not affect the answer

«Ғ» Correct reasoning, but computational errors were made

«Ғ» Only the correct answer is provided or the answer was obtained by considering a specific example

«-» The problem is not solved or is solved incorrectly