kyotsu-test 2014 QCourse1-I-Q1

kyotsu-test · Japan · eju-math__session1 Solving quadratics and applications Determining quadratic function from given conditions
Q1 A quadratic function $y = ax^2 + bx + \frac{3}{a}$ satisfies the following two conditions:
(i) $y$ is maximized at $x = 3$,
(ii) the value of $y$ at $x = 1$ is 2.
We are to find the values of $a$ and $b$.
Using conditions (i) and (ii), we obtain the following relationships between $a$ and $b$:
$$\left\{ \begin{aligned} b & = \mathbf{AB}a \\ \mathbf{C} & = a + b + \frac{\mathbf{D}}{a}. \end{aligned} \right.$$
From these two equalities, we have the equation
$$\mathbf{E}a^2 + \mathbf{F}a - \mathbf{G} = 0$$
and hence
$$a = \mathbf{HI}, \quad b = \mathbf{J}.$$
Thus the maximum value of this function is $\mathbf{K}$.
Q1 A quadratic function $y = ax^2 + bx + \frac{3}{a}$ satisfies the following two conditions:\\
(i) $y$ is maximized at $x = 3$,\\
(ii) the value of $y$ at $x = 1$ is 2.

We are to find the values of $a$ and $b$.

Using conditions (i) and (ii), we obtain the following relationships between $a$ and $b$:

$$\left\{ \begin{aligned}
b & = \mathbf{AB}a \\
\mathbf{C} & = a + b + \frac{\mathbf{D}}{a}.
\end{aligned} \right.$$

From these two equalities, we have the equation

$$\mathbf{E}a^2 + \mathbf{F}a - \mathbf{G} = 0$$

and hence

$$a = \mathbf{HI}, \quad b = \mathbf{J}.$$

Thus the maximum value of this function is $\mathbf{K}$.