Help in QUADRIC

[yorkmanz]

1. In y=a(x−h)^2 +k , what does the value of h tell you about the parabola?

• A. the y -coordinate of the vertex
• B. the x -coordinate of the x -intercept
• C. the x -coordinate of the vertex
• D. the y -coordinate of the y -intercept

my answer was

A ?

CORRECT OR NOT?

1. How does the graph of y=3(x−4)^2 compare to the graph of y=3x^2 ?

• A. It is shifted 4 units up.
• B. It is shifted 4 units down.
• C. It is shifted 4 units to the right.
• D. It is shifted 4 units to the left.

MY ANSWER

C?

CORRECT OR NOT

1. Which of these functions has a maximum y -value of 2?

• A. y=2(x+1)^2 +3
• B. y=3(x+1)^2 +2
• C. y=−3(x+1)^2 +2
• D. y=−(x−2)^2 +3

MY ANSWER IS
C?
CORRECT OR NOT

THANKS MUCH

 

[Deleted member 4993]

1. In y=a(x−h)^2 +k , what does the value of h tell you about the parabola?

• A. the y -coordinate of the vertex
• B. the x -coordinate of the x -intercept
• C. the x -coordinate of the vertex
• D. the y -coordinate of the y -intercept

my answer was

A ?

CORRECT OR NOT? Incorrect. Why did you think that?

1. How does the graph of y=3(x−4)^2 compare to the graph of y=3x^2 ?

• A. It is shifted 4 units up.
• B. It is shifted 4 units down.
• C. It is shifted 4 units to the right.
• D. It is shifted 4 units to the left.

MY ANSWER

C?

CORRECT OR NOT

1. Which of these functions has a maximum y -value of 2?

• A. y=2(x+1)^2 +3
• B. y=3(x+1)^2 +2
• C. y=−3(x+1)^2 +2
• D. y=−(x−2)^2 +3

MY ANSWER IS
C?
CORRECT OR NOT

THANKS MUCH

Thanks for including your answers with the question.

 

[HallsofIvy]

1. In y=a(x−h)^2 +k , what does the value of h tell you about the parabola?

• A. the y -coordinate of the vertex
• B. the x -coordinate of the x -intercept
• C. the x -coordinate of the vertex
• D. the y -coordinate of the y -intercept

my answer was

A ?

CORRECT OR NOT?

No, it's not. If x= h then the y= a(h- h)^2+ k= k. For any other x, since a square is never negative y is larger than k. (h, k) is the vertex. h is the x value of the vertex, not the y value. The correct answer is C.

1. How does the graph of y=3(x−4)^2 compare to the graph of y=3x^2 ?

• A. It is shifted 4 units up.
• B. It is shifted 4 units down.
• C. It is shifted 4 units to the right.
• D. It is shifted 4 units to the left.

MY ANSWER

C?

CORRECT OR NOT

No, that is not correct. y= 3(x- 4)^2 is 0 when x= 4 but is positive for any other x. The graph has vertex at (4, 0). y= 3x^2 is 0 when x= 0 but is positive for any other value of x. The graph has vertex (0, 0). The vertex has been shifted from (4, 0) to (0, 0) so the entire graph has been shifted 4 units to the left. The correct answer is D.

1. Which of these functions has a maximum y -value of 2?

• A. y=2(x+1)^2 +3
• B. y=3(x+1)^2 +2
• C. y=−3(x+1)^2 +2
• D. y=−(x−2)^2 +3

MY ANSWER IS
C?
CORRECT OR NOT

Yes, that is correct! A has vertex at (-1, 3) and opens upward so 3 is the minimum value. B has vertex at (-1, 2) and opens upward so 2 is the minimum value. C has vertex at (-1, 2) and opens downward so 2 is the maximum value. D has vertex at (2, 3) and opens downward so 3 is the minimum value.

THANKS MUCH

 

[yorkmanz]

No, it's not. If x= h then the y= a(h- h)^2+ k= k. For any other x, since a square is never negative y is larger than k. (h, k) is the vertex. h is the x value of the vertex, not the y value. The correct answer is C.



No, that is not correct. y= 3(x- 4)^2 is 0 when x= 4 but is positive for any other x. The graph has vertex at (4, 0). y= 3x^2 is 0 when x= 0 but is positive for any other value of x. The graph has vertex (0, 0). The vertex has been shifted from (4, 0) to (0, 0) so the entire graph has been shifted 4 units to the left. The correct answer is D.







Yes, that is correct! A has vertex at (-1, 3) and opens upward so 3 is the minimum value. B has vertex at (-1, 2) and opens upward so 2 is the minimum value. C has vertex at (-1, 2) and opens downward so 2 is the maximum value. D has vertex at (2, 3) and opens downward so 3 is the minimum value.

Thanks for replying... the SECOND question y= 3(x- 4)^2 the correct answer is ti the right not to the left

Again thanx much