from: category_eng

Euqivalence

A parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$ ?

$mathrm{(A) } 2qquad mathrm{(B) } 5qquad mathrm{(C) } 7qquad mathrm{(D) } 10qquad mathrm{(E) } 11$

1.	Proof

2. 3	Understanding

3. 5	Understanding

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 $a=1$ :

Now the equation is of the form $y=x^2-6x+c$ . Now plug in the point $(2,3)$ and solve for $c$ :

$oxed{c=11 ext{(E)}}$