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Functions - Quadratic Function

For COMPETITION
Number of Total Problems: 4.
FOR PRINT ::: (Book)

Problem Num : 1

From : NCTM

Type: Proof

Section:Functions

Theme:None
Adjustment# :

Difficulty: 3

Category Quadratic Function

Analysis

Solution/Answer

Problem Num : 2

From : NCTM

Type: Understanding

Section:Functions

Theme:None
Adjustment# :

Difficulty: 1

Category Quadratic Function

Analysis

Solution/Answer

Problem Num : 3

From : NCTM

Type: Understanding

Section:Functions

Theme:Euqivalence
Adjustment# :

Difficulty: 1

Category Quadratic Function

Analysis

Solution/Answer

Problem Num : 4

From : AMC10

Type:

Section:Functions

Theme:
Adjustment# : 0

Difficulty: 1

'
A parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$ ?
$mathrm{(A) } 2qquad mathrm{(B) } 5qquad mathrm{(C) } 7qquad mathrm{(D) } 10qquad mathrm{(E) } 11$
'

Category Quadratic Function

Analysis

Solution/Answer
Solution 1

Substitute the points $(2,3)$ and $(4,3)$ into the given equation for $(x,y)$ .
Then we get a system of two equations:
$3=4+2b+c$
$3=16+4b+c$
Subtracting the first equation from the second we have:
$0=12+2b$
$b=-6$
Then using $b=-6$ in the first equation:
$0=1+-12+c$
$c=11 Longrightarrow mathrm{(E)}$ is the answer.

Solution 2

Alternatively, notice that since the equation is that of a monic parabola, the vertex is likely $(3,2)$ . Thus, the form of the equation of the parabola is $y - 2 = (x - 3)^2$ . Expanding this out, we find that $c = 11$ .

Solution 3

The points given have the same $y$ -value, so the vertex lies on the line $x=frac{2+4}{2}=3$ .
The $x$ -coordinate of the vertex is also equal to $frac{-b}{2a}$ , so set this equal to $3$ and solve for $b$ , given that $a=1$ :
$x=frac{-b}{2a}$
$3=frac{-b}{2}$
$6=-b$
$b=-6$
Now the equation is of the form $y=x^2-6x+c$ . Now plug in the point $(2,3)$ and solve for $c$ :
$y=x^2-6x+c$
$3=2^2-6(2)+c$
$3=4-12+c$
$3=-8+c$
$oxed{c=11 ext{(E)}}$

Answer:

Array ( [0] => 3733 [1] => 3745 [2] => 3747 [3] => 7828 ) 4