Question

Which equation represents a graph with a vertex at (1, –6)?

Answer 1

Answer: B) $y = 3x^2 - 6x - 3$

To find vertex (h,k) we use formula

$h=\frac{-b}{2a}$

To find out k we plug in 'h' value in the given equation

Lets find vertex for each equation

A)$y = 3x^2 + 6x - 3$

$h=\frac{-b}{2a}=\frac{-6}{2*3}=-1$

Now plug in -1 for x, $y = 3(-1)^2 + 6(-1) - 3=-6$

So vertex is (-1,-6)

B) $y = 3x^2 - 6x - 3$

$h=\frac{-b}{2a}=\frac{-(-6)}{2*3}=1$

Now plug in -1 for x, $y = 3(1)^2 - 6(1) - 3=-6$

So vertex is (1,-6)

C.y =$3x^2 - 8x - 1$

$h=\frac{-b}{2a}=\frac{-(-8)}{2*3}=\frac{4}{3}$

Now plug in $\frac{4}{3}$  for x, $y = 3(\frac{4}{3})^2 - 8(\frac{4}{3}) - 1= \frac{-19}{3}$

D.$y = 3x^2 - 3x - 6$

$h=\frac{-b}{2a}=\frac{-(-3)}{2*3}= \frac{1}{2}$

Now plug in 1/2 for x,  $y = 3(\frac{1}{2})^2 - 3 (\frac{1}{2})- 6=\frac{-27}{4}$

So, option B has vertex (1,-6)

Answer 2

For the answer to the question above asking Which equation represents a graph with a vertex at (1, –6)?
The answer is the second one among the given choices which is B.y = 3x2 – 6x – 3
For the explanation,
write y+6=3(x^2-1)^2=> 
y+6=3x^2-6x+3=> 
y=3x^2-6x-3