Question

When rounded to the nearest tenth, what are the points of intersection of the system y equals x squared minus three x minus three, y equals x plus one question mark

Graph of the parabolic function y equals x squared minus three times x minus three and the linear function y equals x plus one.

(−0.8, 0.2) and (4.8, 5.8)
(−1, 1) and (5, 6)
(0.2, −0.8) and (6, 5)
(−0.9, 0.1) and (4.9, 5.9)

Answer 1

@whpalmer4 @paki @pooja195 any of you guys think you can help, or know this stuff?

Answer 2

yeah, no problem. what have you done so far?

Answer 3

well, I would like to know how to start this- this kind of math always troubles me, do you have any ideas for me?

Answer 4

yes. is one of the equations easy to solve for one of the variables in terms of the other?

Answer 5

if so, use it to turn two equations in two variables into one equation in one variable, which you then solve...

Answer 6

woah dude slow down... first off I am not yet sure what the unknown variables that are charted on the graph yet.

Answer 7

\[y=x^2-3x-3\]\[y=x+1\]
those are your equations, right?

Answer 8

uh idk you tell me e.e.... i cant really read variables from graphs. however if you say so i believe you. :S

Answer 9

no graph needed, just read the problem statement

Answer 10

oh... i see it now.

Answer 11

ok, so you have two equations, both starting with \(y=\)

Answer 12

like terms first?

Answer 13

take the other side of both equations and set them equal to each other

Answer 14

second equation has a nice, simple expression for the value of \(y\), namely \(x+1\)

so in the first equation, wherever you see \(y\), replace it with \((x+1)\)

Answer 15

that gives you \[(x+1)=x^2-3x-3\]is that something you can solve for the values of \(x\)?

Answer 16

I say values because having an \(x^2\) term as highest power of \(x\) in polynomial means there are 2 solutions, not just 1. we know that from looking at the graph, also, as the line intereects the parabola twice.

Answer 17

intersect* ... but anyways have we yet figured out what the graph reads im still a bit confused?

Answer 18

no need for the graph at all...

Answer 19

but we will check our answers against it when we have them

Answer 20

so, \[x+1=x^2-3x-3\]how are you going to solve that for \(x\)?

Answer 21

how about getting all of the terms on one side, what would that give you?

Answer 22

ok im back had to help my mom with something.

Answer 23

looking at the graph, we can see that one intersection is near \((5,6)\) and the other near \((-1,0)\) but we have two answer choices that are both close to those, and we can't read the graph accurately enough to tell which one is correct.

Answer 24

exactly.

Answer 25

so, we have to use algebra. can you do as I asked with that last equation I posted?

Answer 26

this one "how about getting all of the terms on one side, what would that give you?" or this - 'so,
x+1=x2−3x−3
how are you going to solve that for x?"

Answer 27

whichever, same question :-)

Answer 28

well, can i start by trying to find like terms first? is that an option in this equation?

Answer 29

do whatever you think works...anything is fair game, just do it to both sides of the equation

Answer 30

im really having trouble understanding th whole 'both sides things. Are these opposing sides just to clarify? "x+1 (=) x2−3x−3 <---

Answer 31

if so ... im confused again because im not sure what i can do with that since x + 1 is already in a simple form.

Answer 32

unless, what if i plug in my answers into each equation? My closet estimate would be (−1, 1) and (5, 6)

Answer 33

look, the points where the two intersect are points where they have the same value of \(x\) and the same value of \(y\), right?

Answer 34

so at the spot on the left side where they intersect, the same value of \(x\) plugged into each equation gives exactly the same result for \(y\) from each equation.

that means we can substitute one of the equations into the other.

\[y=x^2-3x-3\]\[y=x+1\]if the left hand side is equal to the right side, and both left hand sides are identical, then we can set the right hand sides equal to each other.

you could of it like this:
\[x^2-3x-3=y=y=x+1\]\[x^2-3x-3=x+1\]now it is just some algebra to solve for \(x\)
\[x^2-3x-3-x=x+1-x\]\[x^2-4x-3=1\]\[x^2-4x-3-1=1-1\]\[x^2-4x-4=0\]
can you solve that equation with the quadratic formula, or factoring the square?