A few days ago
Anonymous

can someone please help with a pre-calc question?

fill out the table for t, x, and y then graph the relation given by: x=3-t^2 and y=1+2t for -2< (or = to) t < (or = to) 2 then theres the t,x,y chart and a graph

Top 5 Answers
A few days ago
    

Favorite Answer

all you need to do is plug in -2 to 2 for t and solve for x and y

t = -2

x = 3 – (-2)^2

x = 3 – 4

x = -1

y = 1 + 2(-2)

y = 1 – 4

y = -3

————————–

t = -1

x = 3 – (-1)^2

x = 3 – 1

x = 2

y = 1 + 2(-1)

y = 1 – 2

y = -1

————————

t = 0

x = 3 – (0)^2

x = 3

y = 1 + 2(0)

y = 1

——————————

t = 1

x = 3 – (1)^2

x = 3 – 1

x = 2

y = 1 + 2(1)

y = 1 + 2

y = 3

———————————

t = 2

x = 3 – (2)^2

x = 3 – 4

x = -1

y = 1 + 2(2)

y = 1 + 4

y = 5

table:

t ……… x ………y

-2 ….. -1 ……. -3

-1 ….. 2 …….. -1

0……. 3 ………1

1 …… 2 ……… 3

2 ……-1 ……… 5

now just plug the points in a graph and connect the dots.

Graph:

the graph is represented by two equations:

y = 1 + 2√(3 – x)

and

y = 1 – 2√(3 – x)

where the domain is x: [-1 , 3]

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A few days ago
Geoff
so first, this is a parametric graph. what you want to do is solve for x and y and then graph as you normally would. they give you the range for t, so plug in values for t to complete the chart.

example, start at t= -2. plug in for the x equation. x=3-(-2)^2=-1. then do it for y. y=1+2(-2)=-1. so your first point is (-1,-1). graph that, and do the same for the rest of the values of t.

0

A few days ago
Anonymous
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A few days ago
Anonymous
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5

A few days ago
‘Old & Cudley’
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