B2. Consider the family of parabolas $y=(k-3) x^{2}-2(k-1) x+2 k+1 ; k \neq 3$.

a) Find the parabola whose axis of symmetry is the line $x=-2$.

b) Find all values of $k$ for which the parabola will have two distinct real roots.
B2.

a) If the axis of symmetry is the line $x=-2$, then $p=-2$. From the equation $p=\frac{-b}{2 a}$, we calculate the desired coefficient $k=\frac{7}{3}$. The equation of the desired parabola is $y=-\frac{2}{3} x^{2}-\frac{8}{3} x+\frac{17}{3}$.

b) For the parabola to have two zeros, the value of the discriminant must be greater than 0. When we substitute the coefficients into the inequality, we get $4(k-1)^{2}-4(k-3)(2 k+1)>0 \Rightarrow k^{2}-3 k-4<0 \Rightarrow$ $(k-4)(k+1)<0$. The solution to the given inequality is $k \in(-1,4)$.

Determining that $p=-2$ 1 point

Calculating the coefficient $k=\frac{7}{3}$ 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=60&width=1739&top_left_y=2454&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=54&width=1739&top_left_y=2509&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=57&width=1739&top_left_y=2556&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=60&width=1736&top_left_y=2603&top_left_x=177)

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=60&width=1722&top_left_y=2654&top_left_x=178)