Graphing Quadratics

[Acewing]

Nevermind, I improperly did the Box and got an entirely different result. There's no common factor here, so the method I was using doesn't work to begin with. Sorry everyone! I got what I was looking for when I tried again.

Hi, it's been a while since I last came here. I'm working on Function Notation, but that requires a grounded understanding of graphing in general, so I've been taking the basic steps, and thus far: I haven't had much trouble with the topic.

That being said, I came across an example that asked for the intercepts of a quadratic equation along with its vertex. Finding the vertex was simple, but when the guide was showing me the steps to find the intercepts, there came a part that I couldn't wrap my head around:

Original Equation - 3x² + x - 2 = y

Setting x = 0 to find the y-intercept leads me too...

3(0)² + 0 - 2 = y

-2 = y, so the y-intercepts are (0, -2)

Now, solving for x to find the y-intercept is where I get confused as to what the guide is doing:

Setting y = 0 to find x-intercept;

3x² + x - 2 = 0

(3x - 2)(x + 1) = 0

3x² - 2 = 0 or x + 1 = 0

x = 2/3 or -1

So x-intercepts would be (2/3, -1)

The bolded part to my understanding, should work out differently, but my train of thought must be wrong, because it is the correct simplification process, right?

To me, (and from what I've reviewed) it should look like:

(3x + 2)(x - 1) = 0

Because --> 3x² + x - 2 = 0

The term without a coefficient; (-2) would be split into its factored numbers: 1 and 2, but they must equal to the central coefficient (+x), otherwise known as (+1) AND has to be capable of being multiplied together back into (-2). So shouldn't it be -1 and +2? Which gives me

(3x + 2)(x - 1) = 0

Rather than (3x - 2)(x + 1) = 0
I believe I'm missing something here, but I'm not sure what, exactly.

PS - I've been juggling between reviewing on PurpleMath and Khan Academy; my brain feels like aquafresh at this point, so I could be forgetting something I've already studied, but it might have slipped my mind entirely, not sure...

 

[Deleted member 4993]

Hi, it's been a while since I last came here. I'm working on Function Notation, but that requires a grounded understanding of graphing in general, so I've been taking the basic steps, and thus far: I haven't had much trouble with the topic.

That being said, I came across an example that asked for the intercepts of a quadratic equation along with its vertex. Finding the vertex was simple, but when the guide was showing me the steps to find the intercepts, there came a part that I couldn't wrap my head around:

Original Equation - 3x² + x - 2 = y ....................You have +x in your original equation

Setting x = 0 to find the y-intercept leads me too...

3(0)² + 0 - 2 = y

-2 = y, so the y-intercepts are (0, -2)

Now, solving for x to find the y-intercept is where I get confused as to what the guide is doing:

Setting y = 0 to find x-intercept;

3x² + x - 2 = 0

(3x - 2)(x + 1) = 0

3x² - 2 = 0 or x + 1 = 0

x = 2/3 or -1

So x-intercepts would be (2/3, -1)

The bolded part to my understanding, should work out differently, but my train of thought must be wrong, because it is the correct simplification process, right?

To me, (and from what I've reviewed) it should look like:

(3x + 2)(x - 1) = 0 .... If you FOIL it you get 3x² - 3x + 2x - 2 = 0 → 3x² - x - 2 = 0 ← not your original equation

Because --> 3x² + x - 2 = 0

The term without a coefficient; (-2) would be split into its factored numbers: 1 and 2, but they must equal to the central coefficient (+x), otherwise known as (+1). So shouldn't it be -1 and +2? Which gives me

(3x + 2)(x - 1) = 0

Rather than (3x - 2)(x + 1) = 0
I believe I'm missing something here, but I'm not sure what, exactly.

PS - I've been juggling between reviewing on PurpleMath and Khan Academy; my brain feels like aquafresh at this point, so I could be forgetting something I've already studied, but it's slipped my mind entirely, not sure...

.

 

[Acewing]

Ahh, indeed, if I were to FOIL and re-construct it into the original equation using (3x + 2)(x - 1), I wouldn't get the original I started with.

Hmm, then in order to solve this dilemma, I should remember to always check to see if the simplified equation fits the original.

It still confuses me, though: if we're to say that -2 and +1 are the two chosen numbers, then it would give me a -1x for the middle coefficient term, when it should be +1x, wouldn't the equation not make sense in that regard?

3x² + x - 2 = 0; 2 = 2 * 1

To get the -2 of the original equation, I need one of those numbers to be a negative, but it also needs to add up to what (+x) is...but if I go with +2 and -1, then (3x + 2)(x - 1) = 3x² - 3x + 2x - 2

Which ultimately leads me to 3x² - x - 2...

It feels like a contradiction, since solving for trinomials for equations with factorable middle and last terms "should" work. It makes even less sense if I were to treat the equation as if it it did not contain any common factors and "box" it.

May someone explain to me what I'm not seeing here, that makes sense? Because to get (3x - 2)(x + 1), that would (in my mind) equate to them using the factors of (-2) and (+1), which adds to (-1)x for the middle term...

Or is it just a matter of always plugging back the zeroes into the original equation to see if it matches up?

 

[Deleted member 4993]

Ahh, indeed, if I were to FOIL and re-construct it into the original equation using (3x + 2)(x - 1), I wouldn't get the original I started with.

Hmm, then in order to solve this dilemma, I should remember to always check to see if the simplified equation fits the original.

It still confuses me, though: if we're to say that -2 and +1 are the two chosen numbers, then it would give me a -1x for the middle coefficient term, when it should be +1x, wouldn't the equation not make sense in that regard?

3x² + x - 2 = 0; 2 = 2 * 1

To get the -2 of the original equation, I need one of those numbers to be a negative, but it also needs to add up to what (+x) is...but if I go with +2 and -1, then (3x + 2)(x - 1) = 3x² - 3x + 2x - 2

Which ultimately leads me to 3x² - x - 2...

It feels like a contradiction, since solving for trinomials for equations with factorable middle and last terms "should" work. It makes even less sense if I were to treat the equation as if it it did not contain any common factors and "box" it.

May someone explain to me what I'm not seeing here, that makes sense? Because to get (3x - 2)(x + 1), that would (in my mind) equate to them using the factors of (-2) and (+1), which adds to (-1)x for the middle term...

Or is it just a matter of always plugging back the zeroes into the original equation to see if it matches up?

If you were to solve

A*x² + B*x + C = 0

You need to look at factors of A*C

In case of 3x² + x -2 = 0, you need to look at factors of -6 [= 3*(-2)] and NOT -2

 

[Dale10101]

Something

Ahh, indeed, if I were to FOIL and re-construct it into the original equation using (3x + 2)(x - 1), I wouldn't get the original I started with.

Hmm, then in order to solve this dilemma, I should remember to always check to see if the simplified equation fits the original.

It still confuses me, though: if we're to say that -2 and +1 are the two chosen numbers, then it would give me a -1x for the middle coefficient term, when it should be +1x, wouldn't the equation not make sense in that regard?

3x² + x - 2 = 0; 2 = 2 * 1

To get the -2 of the original equation, I need one of those numbers to be a negative, but it also needs to add up to what (+x) is...but if I go with +2 and -1, then (3x + 2)(x - 1) = 3x² - 3x + 2x - 2

Which ultimately leads me to 3x² - x - 2...

It feels like a contradiction, since solving for trinomials for equations with factorable middle and last terms "should" work. It makes even less sense if I were to treat the equation as if it it did not contain any common factors and "box" it.

May someone explain to me what I'm not seeing here, that makes sense? Because to get (3x - 2)(x + 1), that would (in my mind) equate to them using the factors of (-2) and (+1), which adds to (-1)x for the middle term...

Or is it just a matter of always plugging back the zeroes into the original equation to see if it matches up?

But the middle term is not determined solely by the two numbers -2 and 1, the middle term is the sum of (-2)(1) +(1)(3) = +1.

The important thing is that factoring quadratics is often a sub procedure in solving more complicated problems and must be done efficiently and mindlessly, that is by procedure. The three common procedures are by "grouping", the "perfect square method" and by using the "quadratic formula". I attached your problem worked by the first two methods.

 

[lookagain]

Original Equation - 3x² + x - 2 = y \(\displaystyle \ \ \ \)

I would recommend using a colon after the word "Equation" instead of the dash.
The dash can be mistaken for a negative sign in this case.

Setting x = 0 to find the y-intercept leads me too...3(0)² + 0 - 2 = y-2 = y, so the y-intercept is (0, -2).

Now, solving for x to find the y-intercept is where I get confused as to what the guide is doing: \(\displaystyle \ \ \ \)You just got finished finding the y-intercept.

Setting y = 0 to find x-intercept(s);

3x² + x - 2 = 0

(3x - 2)(x + 1) = 0

3x² - 2 = 0 \(\displaystyle \ \ \ \) That should just be "3x - 2 = 0."

or x + 1 = 0

x = 2/3 or -1

So x-intercepts would be (2/3, -1) \(\displaystyle \ \ \ \)

No, the x-intercepts are \(\displaystyle \ \)(2/3, 0) \(\displaystyle \ \) and \(\displaystyle \ \) (-1, 0).

.