Respuesta :
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
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) [tex] y = 3x^2 - 6x - 3 [/tex]
To find vertex (h,k) we use formula
[tex] h=\frac{-b}{2a}[/tex]
To find out k we plug in 'h' value in the given equation
Lets find vertex for each equation
A)[tex] y = 3x^2 + 6x - 3 [/tex]
[tex] h=\frac{-b}{2a}=\frac{-6}{2*3}=-1[/tex]
Now plug in -1 for x, [tex] y = 3(-1)^2 + 6(-1) - 3=-6[/tex]
So vertex is (-1,-6)
B) [tex] y = 3x^2 - 6x - 3 [/tex]
[tex] h=\frac{-b}{2a}=\frac{-(-6)}{2*3}=1[/tex]
Now plug in -1 for x, [tex] y = 3(1)^2 - 6(1) - 3=-6[/tex]
So vertex is (1,-6)
C.y =[tex]3x^2 - 8x - 1 [/tex]
[tex]h=\frac{-b}{2a}=\frac{-(-8)}{2*3}=\frac{4}{3}[/tex]
Now plug in [tex]\frac{4}{3}[/tex] for x, [tex]y = 3(\frac{4}{3})^2 - 8(\frac{4}{3}) - 1= \frac{-19}{3}[/tex]
D.[tex] y = 3x^2 - 3x - 6 [/tex]
[tex] h=\frac{-b}{2a}=\frac{-(-3)}{2*3}= \frac{1}{2} [/tex]
Now plug in 1/2 for x, [tex] y = 3(\frac{1}{2})^2 - 3 (\frac{1}{2})- 6=\frac{-27}{4}[/tex]
So, option B has vertex (1,-6)