Asked by jasort20
This is what the directions state to do:
(A) Complete the table,(B)describe the resulting graphs by identifying the vertex point, (c) the graph’s direction, (d) and any axis intercepts gleaned from the table or graph.
Problem #1
Equation is : y = -x + 1
so These are my answers:
(A) Table : (i am going to give it in corrdinate points)
(-1 , 2 )
(0,1)
(1,0)
(B) There is no vertex & no axis of symmetry.
(C) The direction fo line is increasing on the left and decreasing on the right.
(D) Axis of intercepts are : (0,1) and (1,0)
Problem #2 (same directions as one)
Equation: y =x^2 -4x
A)Table:
(-1 , 5 )
(0,0)
(1,-3)
(2,-4)
(3,-3)
(4,0)
(5,5)
(B)Vertex (2,-4)
axis of symmetry x = 2
(c) The direction of the parabola opens upward because the coefficient has a positive x^2.
(d) Axis intercepts are : (0,0) and (4,0)
all correct.
thank you bobpursley
(A) Complete the table,(B)describe the resulting graphs by identifying the vertex point, (c) the graph’s direction, (d) and any axis intercepts gleaned from the table or graph.
Problem #1
Equation is : y = -x + 1
so These are my answers:
(A) Table : (i am going to give it in corrdinate points)
(-1 , 2 )
(0,1)
(1,0)
(B) There is no vertex & no axis of symmetry.
(C) The direction fo line is increasing on the left and decreasing on the right.
(D) Axis of intercepts are : (0,1) and (1,0)
Problem #2 (same directions as one)
Equation: y =x^2 -4x
A)Table:
(-1 , 5 )
(0,0)
(1,-3)
(2,-4)
(3,-3)
(4,0)
(5,5)
(B)Vertex (2,-4)
axis of symmetry x = 2
(c) The direction of the parabola opens upward because the coefficient has a positive x^2.
(d) Axis intercepts are : (0,0) and (4,0)
all correct.
thank you bobpursley
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