Which equation represents a parabola that opens upward, has a minimum value of 3, and has an axis of symmetry at x= 3?
1 answer:
Answer:
Step-by-step explanation:
<u><em>The correct question is</em></u>
Which equation represents a parabola that opens upward, has a minimum value of y=3, and has an axis of symmetry at x= 3?
<em>Verify each case</em>
case 1) we have
This a vertical parabola open upward (the leading coefficient is positive)---> is ok
The vertex represent a minimum ---> is ok
The vertex is the point (-3,-6)
The minimum value of y=-6 ----> is not ok
The axis of symmetry is x=-3 ----> is not ok
case 2) we have
This a vertical parabola open upward (the leading coefficient is positive)---> is ok
The vertex represent a minimum ---> is ok
The vertex is the point (-3,3)
The minimum value of y=3 ----> is ok
The axis of symmetry is x=-3 ----> is not ok
case 3) we have
This a vertical parabola open upward (the leading coefficient is positive)---> is ok
The vertex represent a minimum ---> is ok
The vertex is the point (3,-6)
The minimum value of y=-6 ----> is not ok
The axis of symmetry is x=3 ----> is ok
case 4) we have
This a vertical parabola open upward (the leading coefficient is positive)---> is ok
The vertex represent a minimum ---> is ok
The vertex is the point (3,3)
The minimum value of y=3 ----> is ok
The axis of symmetry is x=3 ----> is ok
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A) There are a number of ways to compute the determinant of a 3x3 matrix. Since k is on the bottom row, it is convenient to compute the cofactors of the numbers on the bottom row. Then the determinant is ...
1×(2×-1 -3×1) -k×(3×-1 -2×1) +2×(3×3 -2×2) = 5 -5k
bi) Π₁ can be written using r = (x, y, z).
Π₁ ⇒ 3x +2y +z = 4
bii) The cross product of the coefficients of λ and μ will give the normal to the plane. The dot-product of that with the constant vector will give the desired constant.
Π₂ ⇒ ((1, 0, 2)×(1, -1, -1))•(x, y, z) = ((1, 0, 2)×(1, -1, -1))•(1, 2, 3)
Π₂ ⇒ 2x +3y -z = 5
c) If the three planes form a sheath, the ranks of their coefficient matrix and that of the augmented matrix must be 2. That is, the determinant must be zero. The value of k that makes the determinant zero is found in part (a) to be -1.
A common approach to determining the rank of a matrix is to reduce it to row echelon form. Then the number of independent rows becomes obvious. (It is the number of non-zero rows.) This form for k=-1 is shown in the picture.
1×(2×-1 -3×1) -k×(3×-1 -2×1) +2×(3×3 -2×2) = 5 -5k
bi) Π₁ can be written using r = (x, y, z).
Π₁ ⇒ 3x +2y +z = 4
bii) The cross product of the coefficients of λ and μ will give the normal to the plane. The dot-product of that with the constant vector will give the desired constant.
Π₂ ⇒ ((1, 0, 2)×(1, -1, -1))•(x, y, z) = ((1, 0, 2)×(1, -1, -1))•(1, 2, 3)
Π₂ ⇒ 2x +3y -z = 5
c) If the three planes form a sheath, the ranks of their coefficient matrix and that of the augmented matrix must be 2. That is, the determinant must be zero. The value of k that makes the determinant zero is found in part (a) to be -1.
A common approach to determining the rank of a matrix is to reduce it to row echelon form. Then the number of independent rows becomes obvious. (It is the number of non-zero rows.) This form for k=-1 is shown in the picture.