Question

Help

Answer 1

let's start with 7 - what are the shapes of the two graphs given?

Answer 2

line and ellipse

Answer 3

well the first equation is an ellipse

Answer 4

good, so that's it for part I

for part II since y = x + 1 we can plug this into equation 1 to get

x^2 + 4(x+1)^2 = 4 solve for your two x values

Answer 5

0,-8/5

Answer 6

good

since y = x + 1 the y-coordinates are just the x-coordinates + 1 (going to do the arithmetic to save some time)

(0,1) and (-8/5,-3/5) as your solutions

Answer 7

anyway, that's part II (just the xvalues) and part III (the x,y, coordinates)

Answer 8

for part 8 I) what shapes are these?

Answer 9

circle and hyperbola

Answer 10

good

for part II simply solve y^2 - x^2 - 9 = 0 for x^2 (not x, solve for x^2)

Answer 11

6?

Answer 12

take this equation

y^2 - x^2 - 9 = 0

add x^2 to both sides, what do you get?

Answer 13

y^2+2x^2-9=0+x^2

Answer 14

-x^2 + x^2 = ?

Answer 15

0

Answer 16

good

so

y^2 - x^2 + x^2 - 9 = x^2

x^2 = y^2 - 9 which is your sol'n for part II

Answer 17

then for part III plug this value of x^2 into the x^2 from equation 1

(y^2-9) +y^2 - 16y + 39 solve for y, should get two numbers

Answer 18

just need to combine like terms

2y^2 - 16y + 30

then factor this

Answer 19

the equation should be equal to 0

Answer 20

3 5

Answer 21

awesome, so those are your y-values for part III

then for part IV plug these back into equation 2 to figure out what the x values would be

Answer 22

5^2-3^2-9=0

Answer 23

3 and 5 are your y-values not your x values

remember when you are solving for an equation there should be one unknown variable, it can't all be numbers

Answer 24

3^2-x^2-9=0
5^2-x^2-9=0

Answer 25

there you go, now solve for x

Answer 26

(4,-4) 0

Answer 27

awesome, so your sol'ns for IV are : (-4,5) (0,3) (4,5)