Question

Find the points of intersection for the given equations

Answer 1

\[x=3-y^2\]

Answer 2

\[y=x-1\]

Answer 3

I tried to use the substitution method but got stuck in the middle of both x and y

Answer 4

the second equation is y = x-1. This means 'y' and 'x-1' are equivalent. We can replace every copy of 'y' with 'x-1' and vice versa

So let's do that in the first equation

\[\Large x = 3 - y^2\]
\[\Large x = 3 - (y)^2\]
\[\Large x = 3 - ({\color{red}{y}})^2\]
\[\Large x = 3 - ({\color{red}{x-1}})^2\]

Hopefully that makes sense so far?

Answer 5

yeah i got a little farther than that when I got stuck

Answer 6

how far did you get?

Answer 7

like i distributed and then didn't know what to do

Answer 8

show me what you got for your last step

Answer 9

x=2-x^2

Answer 10

did you FOIL out (x-1)^2 ?

Answer 11

you need to FOIL before you can distribute

Answer 12

yeah before that i FOILed

Answer 13

Also,
\[\Large (x-1)^2 \ne x^2 + 1^2\]

Answer 14

oh snap i just saw a mistake I made

Answer 15

okay now I have x= -x^2-2x+4

Answer 16

\[\Large x = 3 - (x-1)^2\]
\[\Large x = 3 - (x^2-2x+1)\]
\[\Large x = 3 - x^2+2x-1\]
\[\Large x = -x^2+2x+2\]

Answer 17

Be careful when it comes to distributing. You need to multiply the outer -1 by EVERY term inside

Answer 18

okay I'm not really sure what to do now since it can't be factored

Answer 19

Get everything to one side. Then use the quadratic formula.

Answer 20

so it = 0?

Answer 21

ans then I use the quadratic formula?

Answer 22

yes correct

Answer 23

Tell me what solutions you get

Answer 24

okay I'm stuck again with\[x=\frac{ 3+/-\sqrt{17} }{ -2 }\]

Answer 25

\[\Large x = -x^2+2x+2\]
\[\Large x-x = -x^2+2x+2-x\]
\[\Large 0 = -x^2+x+2\]
\[\Large -x^2+x+2 = 0\]
\[\Large -1*(x^2-x-2) = 0\]
\[\Large x^2-x-2 = 0\]
Hopefully you can see that a = 1, b = -1 and c = -2?

Answer 26

I MADE A REALLY SIMPLE MISTAKE I'M SORRY

Answer 27

that's ok

Answer 28

okay i got x=5 and 4

Answer 29

incorrect

Answer 30

ahh what did I do

Answer 31

what is the value of b^2 - 4ac ?

Answer 32

a = 1, b = -1 and c = -2

b^2 - 4ac = (-1)^2 - 4*(1)*(-2) = ????

Answer 33

7

Answer 34

so x = 4 and 3

Answer 35

(-1)^2 = 1
4*(1)*(-2) = -8

Answer 36

b^2 - 4ac = (-1)^2 - 4*(1)*(-2)
b^2 - 4ac = 1 - (-8)
b^2 - 4ac = 1 + 8
b^2 - 4ac = 9

Answer 37

wait that's what I got before

Answer 38

so x= 1 and 2

Answer 39

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[\Large x = \frac{-(-1)\pm\sqrt{(-1)^2 - 4*(1)*(-2)}}{2(1)}\]
\[\Large x = \frac{1\pm\sqrt{9}}{2}\]
I'll let you finish

Answer 40

`so x= 1 and 2`
one of those is correct

Answer 41

2

Answer 42

\[\Large x = \frac{1\pm\sqrt{9}}{2}\]
\[\Large x = \frac{1\pm 3}{2}\]
\[\Large x = \frac{1+3}{2} \ \ \text{ or } \ \ x = \frac{1-3}{2}\]
\[\Large x = \frac{4}{2} \ \ \text{ or } \ \ x = \frac{-2}{2}\]
\[\Large x = 2 \ \ \text{ or } \ \ x = -1\]

Answer 43

so the solutions to
\[\Large x = 3 - (x-1)^2\]
are x = 2 or x = -1

Answer 44

oh i thought you meant only one of them worked after plugging t back in

Answer 45

if x = 2, then what is the value of y?

Answer 46

1

Answer 47

correct
if x = 2, then y = 1

so one point of intersection is (x,y) = (2,1)

what is the other point of intersection?

Answer 48

(-1,-2)

Answer 49

correct

Answer 50

oh my gosh thank you so much

Answer 51

sorry it took me so long to get this

Answer 52

geogebra confirms these answers

Answer 53

For some reason, geogebra changed \(\Large x = 3-y^2\) into \(\Large y^2+x = 3\)