Question

find the slope of the line containing the indicated pairs of points -3,1 and -6,5

Answer 1

Any ideas?

Answer 2

\[m = \frac{ y _{2}-y _{1} }{ x_{2}-x_{1} }\]

Answer 3

@jkemp are you still there?

Answer 4

yes

Answer 5

What do you know about "slope"?

Answer 6

we are doing them in math and I need help with this one can you help

Answer 7

yes

Answer 8

where do they intersect each other the point that they cross or do they

Answer 9

https://www.youtube.com/watch?v=R948Tsyq4vA

Answer 10

i watched it but this is not answering my question thanks

Answer 11

What @Mandre said is the long and the short of it
\[m = \frac{ y _{2}-y _{1} }{ x_{2}-x_{1} }\]

If that confuses you then may we can think of it conceptually:
Slope is all about differences, specifically, the quotient of differences..
Now, tell me, what is the difference of the two y-values of the points given?

Answer 12

^Just subtract them.

Answer 13

ok i got it thanks

Answer 14

What did you get for your answer?

Answer 15

where can I watch videos on solution sets using graphs

Answer 16

No idea... @skullpatrol ?

Answer 17

how about binomials

Answer 18

What about them? :)

Answer 19

or quadratic formula ot factoring

Answer 20

videos on how to work them my Professor had to speed bump thru these sections I am 53 and never took this in school so I am lost and finals are next week

Answer 21

Oh, wouldn't you know it, I wrote a tutorial about that a while back :3

Answer 22

Or actually, more like deriving...

Answer 23

To be general when it comes to solving quadratic equations of the form
\[\Large \color{red}ax^2 + \color{blue}bx + \color{green}c =0\]
We will use the method of 'completing the square'
First, we need to have the \(x^2\) alone, with no coefficient, so the first step is to divide everything by \(\color{red}a\)
\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{green}c}{\color{red}a}=0\]
Next, we subtract \(\Large \frac{\color{green}c}{\color{red}a}\) from both sides to that all the terms without \(x\) go on the right side...
\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x=-\frac{\color{green}c}{\color{red}a}\]


Now, to the actual completing of the square... recall that to complete the square given
\[\Large x^2 + \color{purple}px= k\]we take half of the coefficient of the lone \(x\) (the one with no exponent), in this case, \(\color{purple}p\), giving \(\Large \frac{\color{purple}p}{2}\) and then square it, yielding \(\Large \frac{\color{purple}p^2}{4}\) and add THAT to both sides of the equation...\[\Large x^2 +\color{purple}px+\frac{\color{purple}p^2}{4}=k+\frac{\color{purple}p^2}{4}\]such that the left side is now a perfect square...\[\Large \left(x+\frac{\color{purple}p}2\right)^2 = k+\frac{\color{purple}p^2}{4}\]


It so happens that in this case, our \(\color{purple}p\) value is \(\Large \frac{\color{blue}b}{\color{red}a}\). So half of that is \(\Large \frac{\color{blue}b}{2\color{red}a}\). Squaring this yields \(\LARGE \frac{\color{blue}b^2}{4\color{red}a^2}\) And this is now added to both sides of the equation \(\Large x^2 +\frac{\color{blue}b}{\color{red}a}x=-\frac{\color{green}c}{\color{red}a}\) giving us...

\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{blue}b^2}{4\color{red}a^2}=\frac{\color{blue}b^2}{4\color{red}a^2}-\frac{\color{green}c}{\color{red}a}\]

At this point, we may want to simplify the right-hand side of the equation into a single fraction, this can be done using their LCD, which happens to be \(\large 4\color{red}a^2 \)

\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{blue}b^2}{4\color{red}a^2}=\frac{\color{blue}b^2-4\color{red}a\color{green}c}{4\color{red}a^2}\]


Now, as intended, the left-hand side is now a perfect square, so...
\[\Large \left(x+\frac{\color{blue}b}{2\color{red}a}\right)^2=\frac{\color{blue}b^2-4\color{red}a\color{green}c}{4\color{red}a^2}\]

Taking the square root of both sides yields...\[\Large x+\frac{\color{blue}b}{2\color{red}a}=\pm\sqrt{\frac{\color{blue}b^2-4\color{red}a\color{green}c}{4\color{red}a^2}}\]
Simplifying...\[\Large x+\frac{\color{blue}b}{2\color{red}a}=\frac{\pm\sqrt{\color{blue}b^2-4\color{red}a\color{green}c}}{2\color{red}a}\]

Bringing the term \(\Large \frac{\color{blue}b}{2\color{red}a}\) to the right-hand side by subtracting from both sides gives
\[\Large x=-\frac{\color{blue}b}{2\color{red}a}\pm \frac{\sqrt{\color{blue}b^2-4\color{red}a\color{green}c}}{2\color{red}a}\]

And finally, combining, since they have the same denominator, we get...
\[\Large x= \frac{-\color{blue}b\pm \sqrt{\color{blue}b^2-4\color{red}a\color{green}c}}{2\color{red}a}\]


And this is the ever-notorious quadratic formula ^_^

Answer 24

ok I will try to solve one to see if I get it I got the slope thanks you should teach

Answer 25

I'll certainly think about it :)
Give me a holler if you need any more help (and I'm online ^_^)

Answer 26

ok this is to confusing I need to see what is happening is there a video

Answer 27

I understand... I don't know if there is a video, but if you like, I can walk you through the steps of solving a quadratic equation :)

Answer 28

ok i have 3x2 - 16x - 12=0 the two is squard

Answer 29

And you want to solve this using the quadratic formula? :)

Answer 30

yes or factoring which ever one is best

Answer 31

Well, factoring, can it be factored? (Sorry about that, factoring isn't one of the things I can actually explain... better outsource that particular bit of knowledge :) )

Answer 32

ok then quadratic formula

Answer 33

Okay, apparently, it cannot be factored :3
Oh well...
Standard quadratic equation is
\[\Large \color{red}ax^2 + \color{blue}bx + \color{green}c =0\]
Right?

Answer 34

which one is a the 3 and b is the 16

Answer 35

c is my constant

Answer 36

c is 12.
But be careful, your b is -16, not simply 16.

Pay attention to signs... one wrong sign could very well mean a completely wrong answer ^_^

Answer 37

ok if both signs in the equations are negative i would add but still use a negative 16 right

Answer 38

I'm afraid I don't get what you mean :)
Gist of the matter is that
\[\Large \color{red}3x^2\color{blue}{-16}x \color{green}{+12}=0\]
Your a and c value are positive and your b-value is negative ^_^

Answer 39

ok i change the subtraction sign at the end to positive what about between the 3x nd 16x

Answer 40

What subtraction sign at the end? D:

Answer 41

3x2-16x-12=0

Answer 42

Oh, right, sorry, my bad, c is -12

Answer 43

and what if you have a 3x2 +20x-7=0 what would I do here

Answer 44

Just keep the signs :)

Answer 45

Sorry, I was wrong -.-

Answer 46

\[\Large \color{red}3x^2\color{blue}{-16}x \color{green}{-12}=0\]
My apologies, I was the one that misread the signs :3

Answer 47

2 negative make a positive , what about a positive then a negitive

Answer 48

ok just keep all negative

Answer 49

except the 3 of course :)

Answer 50

So, with this \[\Large \color{red}3x^2\color{blue}{-16}x \color{green}{-12}=0\]

You may now use the quadratic formula:
\[\Large x = \frac{-\color{blue}b \pm \sqrt{\color{blue}b^2 - 4\color{red}a\color{green}c}}{2\color{red}a}\]