11.39 Find the coefficients of the quadratic function $y=a x^{2}+b x+c$, knowing that at $x=-0.75$ it takes the maximum value of 3.25, and at $x=0$ it takes the value 1.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-342.jpg?height=325&width=282&top_left_y=549&top_left_x=857)

Fig. 11.1
11.39 According to the condition, when $x=0$, the value of $y=1$, from which $c=1$. Further, since the abscissa of the vertex of the parabola $x_{\mathrm{s}}=-\frac{b}{2 a}$ and at $x=-0.75$ we have $y_{\text {max }}=3.25$, then $-\frac{b}{2 a}=-\frac{3}{4}$, from which $b=\frac{3}{2} a$. Then the desired function will take the form $y=a x^{2}+\frac{3}{2} a x+1$. Substituting the coordinates of the vertex of the parabola into this equality, we get $\frac{13}{4}=\frac{9}{16} a+\frac{3}{2} a\left(-\frac{3}{4}\right)+1$. Solving the last equation with respect to $a$ and finding $a=-4$; hence, $b=-6$. Thus, the desired function has the form $y=-4 x^{2}-6 x+1$.