If the vertex of the parabola
$y=3x^2-12x+9$ is $(a,b)$
Then the parabola(s) whose vertex is $(b,a)$ is (are)
A) $y=x^2+6x+11$
B) $y=x^2-7x+3$
C) $y=-2x^2-12x-16$
D) $y=-2x^2+16x-13$
This is a multiple choice based question. I tried, but I am not able to get it. Please help me out.
Best Answer
Write your function as : $y=3(x^2-4x+3)=3(x-3)(x-1)$. The axis of symmetry (x coordinate of vertex) is $\frac{3+1}{2}=2$.Then plug in to find $y$ value of vertex which is $-3$.
So $(a,b)=(2,-3)$
Now find vertex of all your options (axis of symmetry can be calculated by $\frac{-b}{2a}$):
A)
$\frac{-6}{2}$ which is $-3$ So check $y$ value and see that it is $2$ which is $(b,a)$ as desired.
B)
$\frac{7}{2}$ not right either
C)
$\frac{-12}{4}=-3$ Plug in to find $y$ value of vertex and see that it is $2$.
D)
$\frac{16}{4}$ not right either
So A and C are the correct answers