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MissPhiladelphia

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

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)

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