## Task B-2.7.

The graph of the quadratic function $f(x)=a x^{2}+b x+c$ has its vertex at the point $(t, t)$ and passes through the point $(-t,-t)$. Determine all values of the real number $t, t \neq 0$ such that $a^{2}+b^{2}+c^{2}=\frac{33}{16}$.
## Solution.

The abscissa of the vertex of a quadratic function is $t=-\frac{b}{2 a}$, so $b=-2 a t$.

Substitute the coordinates of the points $(t, t)$ and $(-t,-t)$ into the graph equation. Then we have $t=a t^{2}+b t+c$, or $t=-a t^{2}+c$ (because $\left.b=-2 a t\right)$ and

$-t=a t^{2}-b t+c$ or $t=-3 a t^{2}-c$.

It follows that $4 a t^{2}=-2 t$, so $a=-\frac{1}{2 t}$.

From this, we get $b=-2 a t=1$, and $c=t+a t^{2}=\frac{t}{2}$.

After substituting the obtained expressions for $a, b$, and $c$ into the equality $a^{2}+b^{2}+c^{2}=\frac{33}{16}$, we get the equation $\frac{1}{4 t^{2}}+1+\frac{t^{2}}{4}=\frac{33}{16}$,

which simplifies to the biquadratic equation $4 t^{4}-17 t^{2}+4=0$.

Then $t^{2}=4, t^{2}=\frac{1}{4}$, or

$t \in\left\{-2,-\frac{1}{2}, \frac{1}{2}, 2\right\}$.

1 point

## Note:

The student can obtain the coefficients $a, b$, and $c$ in the following ways (for which they receive the marked points):

1. method:

The abscissa of the vertex of a quadratic function is $t=-\frac{b}{2 a}$, so $b=-2 a t$.

The ordinate of the vertex of a quadratic function is $t=\frac{4 a c-b^{2}}{4 a}$, from which, with $b=-2 a t$, we get $c=a t^{2}+t$.

Substituting the coordinates of the point $(-t,-t)$ into the graph equation, we get $-t=a t^{2}-b t+c$, from which, with $b=-2 a t$ and $c=a t^{2}+t$, it follows that

$-t=a t^{2}+2 a t^{2}+a t^{2}+t$, or $a=-\frac{1}{2 t}$

$b=-2 a t=1$, and $c=t+a t^{2}=\frac{t}{2}$.

2 points

2. method:

The function $f$ can be written in the form $f(x)=a\left(x-x_{0}\right)^{2}+y_{0}$, where $\left(x_{0}, y_{0}\right)$ is the vertex of the graph of the function. Then $f(x)=a(x-t)^{2}+t$.

Substituting the coordinates of the point $(-t,-t)$ into the graph equation, we get $-t=4 a t^{2}+t$, or $a=-\frac{1}{2 t}$.

Therefore, $f(x)=-\frac{1}{2 t} x^{2}+x+\frac{t}{2}$, so $b=1, c=\frac{t}{2}$.

3. method:

Since the point $(t, t)$ is the vertex of the graph of the quadratic function, and the point $(-t,-t)$ belongs to the graph of this function, then due to symmetry, the point $(3 t,-t)$ also belongs to the graph of this function.

Substituting the coordinates of these points into $f(x)=a x^{2}+b x+c$, we get the system

$a t^{2}+b t+c=t$

$a t^{2}-b t+c=-t$

$9 a t^{2}+3 b t+c=-t$

Multiplying the first equation by -1 and adding it to the second, we get $2 b t=2 t \Rightarrow b=1$ $(t \neq 0)$.

1 point

Substituting $b=1$ into the second and third equations, we get

$a t^{2}+c=0$

$9 a t^{2}+c=-4 t$.

From this, we get: $a=-\frac{1}{2 t}$ and $c=\frac{t}{2}$.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

3rd grade - high school - B variant
February 4, 2021.

## IF THE STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.