Parametric equation

Eliminate the parameter. Convert the parametric equation to rectangular.

x= 2+4t and y=-1+6t

• y=(3/2)x - 4

• t=(x-2)/4

• y=x - 4

• y= (2/3)x + 4

Tags

CCSS.HSA.CED.A.2

CCSS.HSA.REI.D.10

CCSS.HSA.REI.C.6

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Which set of parametric equations satisfies the given data table?

• x(t) = t³ - 1
  y(t) = 2t
• x(t) = t³ - 1
  y(t) = t + 2
• x(t) = t + 8
  y(t) = t + 2
• x(t) = t + 8
  y(t) = 2t

Tags

CCSS.HSA.CED.A.2

CCSS.HSA.REI.D.10

What is the "starting point" of this parametric function? (t=0)

• (1, -1)

• (4, -3)

• (-3, 2)

• (-3, 4)

Tags

CCSS.HSA.CED.A.2

CCSS.HSA.REI.D.10

CCSS.HSA.REI.C.6

CCSS.HSA.SSE.A.1

Calculate a rectangular equation (y=mx+b) for the parametric function
x = -2 + 2t
y = 2 - 1t

• y = -x +1
• y = -x + 2
• y = 1/2x +1
• y = -1/2x + 1

Tags

CCSS.HSA.REI.C.6

CCSS.8.EE.C.8B

Sketch the curve of x(t) = 2t + 1 and y(t) = t² + 2.

• A
• B
• C
• D

Tags

CCSS.HSA.CED.A.2

CCSS.HSA.REI.D.10

CCSS.HSF.IF.A.1

CCSS.HSF.IF.B.4

CCSS.HSF.IF.C.7

Identify the starting point of the parametric equation.
x = -2 + 2t
y = 2 - 1t

• (-2, 2)
• (2, -2)
• (2, -1)
• (-1, 2)

Tags

CCSS.HSA.REI.C.6

CCSS.HSA.REI.C.7

A curve is given parametrically by the equations, x = t and y = t² + 1. Find dy/dx in terms of t

• t²

• 2t²

• 1/2t

• 2t

A curve is given parametrically by the equations, x = t and y = t² + 1. Find the equation of the tangent at the point A (1,1) on the curve. Given your answer in the form y = mx + c

• y = x -2

• y = 2x +1

• y = 2x -1

• y = x + 2

Tags

CCSS.HSA.REI.C.7

Eliminate the parameter. 
x = t
y = 5t

• y=5x²
• y=5x
• y= x - 5
• y=x²

Tags

CCSS.HSA.REI.C.8

Convert from parametric to rectangular:  x = t - 4  and y = 2t + 1
Hint:  Solve for t and then substitute into the "y" equation.

• y = 2t - 4
• y = 2x + 9
• y = 2x + 4
• y = 2t - 3

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3