Determining quadratic function from given conditions

The question specifies conditions on a quadratic function (vertex, specific values, maximum/minimum, passing through points) and asks to determine its coefficients, expression, or extremal value.

csat-suneung 2017 Q18 4 marks View
A quadratic function $f ( x )$ with leading coefficient 1 satisfies $$\lim _ { x \rightarrow a } \frac { f ( x ) - ( x - a ) } { f ( x ) + ( x - a ) } = \frac { 3 } { 5 }$$ When the two roots of the equation $f ( x ) = 0$ are $\alpha$ and $\beta$, what is the value of $| \alpha - \beta |$? (Here, $a$ is a constant.) [4 points]
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5
gaokao 2015 Q20 15 marks View
20. (15 points) Let the function $f ( x ) = x ^ { 2 } + a x + b , ( a , b \in R )$ .
(1) When $b = \frac { a ^ { 2 } } { 4 } + 1$ , find the expression for the minimum value $g ( a )$ of the function $f ( x )$ on $[ - 1,1 ]$ ;
(2) Given that the function $f ( x )$ has a zero on $[ - 1,1 ]$ , $0 \leq b - 2 a \leq 1$ , find the range of values for $b$ .
jee-main 2012 Q61 View
If $a , b , c \in \mathrm { R }$ and 1 is a root of equation $a x ^ { 2 } + b x + c = 0$, then the curve $y = 4 a x ^ { 2 } + 3 b x + 2 c , a \neq 0$ intersect $x$-axis at
(1) two distinct points whose coordinates are always rational numbers
(2) no point
(3) exactly two distinct points
(4) exactly one point
jee-main 2020 Q51 View
Let $f ( x )$ be a quadratic polynomial such that $f ( - 1 ) + f ( 2 ) = 0$. If one of the roots of $f ( x ) = 0$ is 3 , then its other root lies in
(1) $( - 1,0 )$
(2) $( 1,3 )$
(3) $( - 3 , - 1 )$
(4) $( 0,1 )$
jee-main 2021 Q74 View
Let $P ( x ) = x ^ { 2 } + bx + c$ be a quadratic polynomial with real coefficients such that $\int _ { 0 } ^ { 1 } P ( x ) d x = 1$ and $P ( x )$ leaves remainder 5 when it is divided by $( x - 2 )$. Then the value of $9 ( b + c )$ is equal to:
(1) 9
(2) 15
(3) 7
(4) 11
jee-main 2022 Q68 View
Let $f ( x ) = a x ^ { 2 } + b x + c$ be such that $f ( 1 ) = 3 , f ( - 2 ) = \lambda$ and $f ( 3 ) = 4$. If $f ( 0 ) + f ( 1 ) + f ( - 2 ) + f ( 3 ) = 14$, then $\lambda$ is equal to
(1) $- 4$
(2) $\frac { 13 } { 2 }$
(3) $\frac { 23 } { 2 }$
(4) $4$
jee-main 2025 Q2 View
Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a function defined by $f ( x ) = ( 2 + 3 a ) x ^ { 2 } + \left( \frac { a + 2 } { a - 1 } \right) x + b , a \neq 1$. If $f ( x + \mathrm { y } ) = f ( x ) + f ( \mathrm { y } ) + 1 - \frac { 2 } { 7 } x \mathrm { y }$, then the value of $28 \sum _ { i = 1 } ^ { 5 } | f ( i ) |$ is
(1) 545
(2) 715
(3) 735
(4) 675
kyotsu-test 2014 QCourse1-I-Q1 View
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 $\mathbf{K}$.
kyotsu-test 2014 QCourse2-I-Q1 View
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 $\mathbf{K}$.
kyotsu-test 2014 QCourse1-I-Q1 View
Let $a$ and $b$ be real numbers, where $a > 0$. Consider the two quadratic functions
$$f ( x ) = 2 x ^ { 2 } - 4 x + 5 , \quad g ( x ) = x ^ { 2 } + a x + b .$$
We are to find the values of $a$ and $b$ when the function $g ( x )$ satisfies the following two conditions.
(i) The minimum value of $g ( x )$ is 8 less than the minimum value of $f ( x )$.
(ii) There exists only one $x$ which satisfies $f ( x ) = g ( x )$.
Since the minimum value of $f ( x )$ is $\mathbf { A }$, from condition (i), we derive the equality
$$b = \frac { a ^ { 2 } } { \mathbf { B } } - \mathbf { C } \text {. }$$
Hence the equation from which we can find the $x$ satisfying $f ( x ) = g ( x )$ is
$$x ^ { 2 } - ( a + \mathbf { D } ) x - \frac { a ^ { 2 } } { \mathbf { E } } + \mathbf { F G } = 0 .$$
Thus, since $a > 0$, from condition (ii) we obtain
$$a = \mathbf { H } , \quad b = \mathbf { I J } .$$
In this case, the $x$ satisfying $f ( x ) = g ( x )$ is $\square \mathbf{ K }$.
kyotsu-test 2014 QCourse2-I-Q1 View
Let $a$ and $b$ be real numbers, where $a > 0$. Consider the two quadratic functions
$$f ( x ) = 2 x ^ { 2 } - 4 x + 5 , \quad g ( x ) = x ^ { 2 } + a x + b .$$
We are to find the values of $a$ and $b$ when the function $g ( x )$ satisfies the following two conditions.
(i) The minimum value of $g ( x )$ is 8 less than the minimum value of $f ( x )$.
(ii) There exists only one $x$ which satisfies $f ( x ) = g ( x )$.
Since the minimum value of $f ( x )$ is $\mathbf{A}$, from condition (i), we derive the equality
$$b = \frac { a ^ { 2 } } { \mathbf { B } } - \mathbf { C } \text {. }$$
Hence the equation from which we can find the $x$ satisfying $f ( x ) = g ( x )$ is
$$x ^ { 2 } - ( a + \mathbf { D } ) x - \frac { a ^ { 2 } } { \mathbf { E } } + \mathbf { F G } = 0 .$$
Thus, since $a > 0$, from condition (ii) we obtain
$$a = \mathbf { H } , \quad b = \mathbf { I J } .$$
In this case, the $x$ satisfying $f ( x ) = g ( x )$ is $\mathbf { K }$.
kyotsu-test 2015 QCourse1-II-Q1 View
Consider the two parabolas $$\begin{aligned} \ell : & & y = ax^2 + 2bx + c \\ m : & & y = (a+1)x^2 + 2(b+2)x + c + 3. \end{aligned}$$ Four points A, B, C and D are assumed to be in the relative positions shown in the figure. One of the two parabolas passes through the three points A, B and C, and the other one passes through the three points B, C and D.
(1) The parabola passing through the three points A, B and C is $\mathbf{A}$. Here, for $\mathbf{A}$ choose the correct answer from (0) or (1), just below. (0) parabola $\ell$ (1) parabola $m$
(2) Since both parabolas $\ell$ and $m$ pass through the two points B and C, the $x$-coordinates of B and C are the solutions of the quadratic equation $$x^2 + \mathbf{B}x + \mathbf{C} = 0.$$ Hence, the $x$-coordinate of point B is $\mathbf{DE}$, and the $x$-coordinate of point C is $\mathbf{FG}$.
(3) In particular, we are to find the values of $a$, $b$ and $c$ when $\mathrm{AB} = \mathrm{BC}$ and $\mathrm{CO} = \mathrm{OD}$.
Since the two points C and D are symmetric with respect to the $y$-axis, we have $b = \mathbf{H}$. On the other hand, since $\mathrm{AB} = \mathrm{BC}$, the straight line $x = \mathbf{IJ}$ is the axis of symmetry of $\mathbf{A}$. Hence we have $a = -\frac{\mathbf{K}}{\mathbf{L}}$. And we have $c = \frac{\mathbf{M}}{\mathbf{L}}$.
kyotsu-test 2018 QCourse1-II-Q2 View
Let $a$ be a constant other than 0 . Let
$$\begin{aligned} & f ( x ) = x ^ { 2 } + 2 a x - 4 a - 12 , \\ & g ( x ) = a x ^ { 2 } + 2 x - 4 a + 4 . \end{aligned}$$
(1) When the solutions of $f ( x ) = 0$ and the solutions of $g ( x ) = 0$ coincide, $a$ is $\mathbf { M N }$, and their solutions are $x = \mathbf { O P }$ and $x = \mathbf { Q }$.
(2) $g ( x ) = 0$ has just one solution when $a = \frac { \mathbf { R } } { \mathbf { S } }$, and in this case the solution is $x =$ $\mathbf{TU}$.
(3) The range of $a$ such that $f ( x ) < g ( x )$ for all $x$ is $\mathbf{VW}$.