SOLUTION: I need someone help i am stuck doing this last part of my homework. can someone explain it to me
Match each equation for a parabola with the direction that the parabola opens
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-> SOLUTION: I need someone help i am stuck doing this last part of my homework. can someone explain it to me
Match each equation for a parabola with the direction that the parabola opens
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Question 623928: I need someone help i am stuck doing this last part of my homework. can someone explain it to me
Match each equation for a parabola with the direction that the parabola opens.
1. y=-3(x+1)^2-8
2. x=-3(y+1)^2-8
3. y=3(x+1)^2-8
4. x=3(y+1)^2-8
A. up
B. right
C. down
D. left
Hi
Note:
the vertex form of a Parabola opening Up(a>0) or Down(a<0) is where(h,k) is the vertex.
the vertex form of a Parabola opening Right(a>0) or Left(a<0) is where(h,k) is the vertex.
1. y=-3(x+1)^2-8 Down a = -3 < 0
2. x=-3(y+1)^2-8 Left a = -3 < 0
3. y=3(x+1)^2-8 Up a = 3 > 0
4. x=3(y+1)^2-8 Right a = 3 > 0
You can put this solution on YOUR website! if y = x^2, then the parabola will open up.
if y = -x^2, then the parabole will open down.
if x = y^2, then the parabola will open to the right.
if x = -y^2, then the parabole will open to the left.
what we are dealing with is the coefficient of the squared term.
your equations are:
1. y=-3(x+1)^2-8
2. x=-3(y+1)^2-8
3. y=3(x+1)^2-8
4. x=3(y+1)^2-8
(x+1)^2 will be equal to x^2 + 2x + 1
(y+1)^2 will be equal to y^2 + 2y + 1
your equations therefore become:
1. y = -3(x^2 + 2x + 1) - 8
2. x = -3(y^2 + 2y + 1) - 8
3. y = 3(x^2 + 2x + 1) - 8
4. x = 3(y^2 + 2y + 1) - 8
remove parentheses and simplify to get:
1. y = -3x^2 - 6x - 11
2. x = -3y^2 - 6y - 11
3. y = 3x^2 + 6x - 5
4. x = 3y^2 + 3y - 5
when determining whether the parabole will open up or down, the only meaningful consideration is the sign of the coefficient of the squared term.
based on this, you should get the following answers.
1. will open down
2. will open to the left.
3. will open up.
4. will open to the right.
we can graph these equations and see how we did.
to graph these equations, you have to solve for y in all equations.
the original equations are:
1. y=-3(x+1)^2-8
2. x=-3(y+1)^2-8
3. y=3(x+1)^2-8
4. x=3(y+1)^2-8
equations 1 and 3 can be graphed as is.
you need to solve for y in equations 2 and 4.
equation 2 is:
x = -3(y+1)^2 - 8
add 3(y+1)^2 to both sides of this equation and subtract x from both sides of this equation to get:
3(y+1)^2 = -x - 8
divide both sides of this equation by 3 to get:
(y+1)^2 = (-x-8)/3 which is the same as:
(y+1)^2 = -(x+8)/3
take the square root of both sides of this equation to get:
y+1 = +/- ( -(x+8)/3 )
subtract 1 from both sides of this equation to get:
y = -1 +/- sqrt( -(x+8)/3 )
equations to graph for equation 2 are:
y = -1 + sqrt( -(x+8)/3 )
y = -1 - sqrt( -(x+8)/3 )
equation 4 is:
x = 3(y+1)^2 - 8
add 8 to both sides of this equation to get:
3(y+1)^2 = x + 8
divide both sides of this equation by 3 to get:
(y+1)^2 = (x+8)/3
take the square root of both sides of this equation to get:
y+1 = +/- sqrt( (x+8)/3 )
subtract 1 from both sides of this equation to get:
y = -1 +/- sqrt( (x+8)/3 )
equations to graph for equation 4 are:
y = -1 + sqrt( (x+8)/3 )
y = -1 - sqrt( (x+8)/3 )
graph of equation 1 is:
graph of equation 2 is:
graph of equation 3 is:
graph of equation 4 is:
