Graphing quadratic equations.

One of the X intercepts was wrong. It should be -5 and +2 not -2.
(x+5)=0 and (x-2)=0 so x=-5 and x=2

 

Haha oops I screwed that up... Hehe been quit a while since I had to do any of this, but thats the general idea...

 

from these three point the two x intercepts x=-5 and 2 and then the y intercept y=-10 you will be able to make the graph. This one is a parabola that is positive. The cup will be able to hold water.

 

This is what I meant ---> (x+5)(x-2)=(x^2+3x-10), I wasn't paying too much attention... :&
Anyway that's just the format I'm used to...

 

yeah I am having to recall everything I learned in Calculus last year. It is slowly coming back to me though

 

Yeah I have the general idea how to do most of it, just not the specifics... A bit of practice and I'd be back up to speed... I hope so anyway...

 

at least it is not integrals or differentiating

 

hahaha... math fucking sucks!!! i'm going to have to take algebra one B fer the fourth time, prolly next semester. i never could get these... or anything else

 

You said once that only three people in your class were passing and that your teacher always yells at you if you don't remember how to do the problem... that's how my teacher was last year, but he had a reason - nobody would listen during the first half of the class... we'd always have really good days, then other days where he snapped... lol, but he was a really good guy. At the end of the year, we did nothing for a month after we took our tests... that's the only good thing about Standards Of Learning and standardized testing... once you do the end of the year tests, you don't do much. But I feel that homeschooling beats it any day.

 

I have a feeling that I'm a bit late on this thread. But maybe some of this will help for future reference. I'm probably re-hashing some of what has been already said, but this is how I'd go about doing the original problem that you asked about.

You were given this equation:
y = 9x^2 - 36x + 35

Looking at it, you know that the graph is a parabola because the highest exponent in the equation is 2 (9x^2), and of course the graph for a quadratic equation is a parabola. You also know that the parabola opens upward because the coefficient of the quadratic term (9x^2) is positive (because 9 > 0). If the quadratic term was -9x^2 instead of 9x^2, it would open downward (because -9 < 0).

The vertex is the lowest point on the parabola if it opens upward (like the one in this problem does). Conversely, the vertex would be the highest point on the parabola if it opened downward.

The equation for the vertex of a parabola is x = -b / 2a. The quadratic term is the "A" term (9x^2) and linear term is the "B" term (-36x). So the vertex of this upward-opening parabola is:
x = -(-36) / 2(9)
x = 2

So far, we have the x-coordinate value of the vertex (2, y). Now to get its y-coordinate, substitute "2" into the original equation for x.
y = 9(2)^2 - 36(2) + 35
y = -1

Therefore, the vertex is located at point (2, -1). Remember that this will be the lowest point on the parabola when you are graphing it.


To find the y-intercept (point at which the parabola crosses the y-axis), substitute "0" into the original equation for x.
y = 9(0)^2 - 36(0) + 35
y = 35

Therefore, the y-intercept is located at point (0, 35).


To find the x-intercepts (points at which the parabola crosses the x-axis), set the equation equal to 0.
9x^2 - 36x + 35 = 0

Factor it:
(3x - 7)(3x - 5) = 0

By the zero product rule, one of these two equations [either (3X - 7) or (3x - 5)] HAVE to equal 0 in order for their product to be 0.

Break apart the two equations and set them equal to 0.
3x - 7 = 0
3x = 7
x = 7/3

You must check to verify if this value (7/3) works in the original equation because the equation that you just solved for may NOT work (since only ONE of the equations are actually required to equal 0 under the zero product rule).

To check it, substitute "7/3" in for x and set it equal to 0.
9(7/3)^2 - 36(7/3) + 35 = 0

After you do a bunch of cross-multiplying and stuff, you end up with 0 = 0. This is true of course, so you know that 7/3 works. We still have to check the other equation from before.
3x - 5 = 0
3x = 5
x = 5/3

Again, substitute "5/3" into the original equation:
9(5/3)^2 - 36(5/3) + 35 = 0

Calculate the left side:
0 = 0 (True again)

Both 7/3 and 5/3 satisfy this particular equation, so the x-intercepts will be (0, 7/3) and (0, 5/3). It sucks that we went through all that work to verify those values, but sometimes one of them may not work.

Draw a graph and plot the x-intercepts [(7/3, 0), (5/3, 0)], y-intercept [(0, 35)], and vertex [(2, -1)]. The problem tells you to plot additional points (where x is between -10 and 10) so that it's easier for you to graph it. So if we pick where x = 1 (and substitute it into the original equation), we'll get:
y = 9(1)^2 - 36(1) + 35
y = 8

Therefore, the parabola will pass through point (1, 8). You can pick other points if you want, but for this particular problem, beware that the y-coordinates increase very, very quickly.

As long as you know the general shape and behavior of a parabola, you can graph it somewhat easily. If you were to draw a vertical line through the vertex, it would create a line of symmetry. So you really only have to know what the graph looks like on one side of the line of symmetry in order to plot it because it will be perfectly symmetrical. More points ensure greater accuracy, though.

*Phew* I hope some of that helped.

 

well, we listen as goode as we can, but it's sooo boring and he makes it really hard. all he does is write it on the board. no explaining or anything. other people have complained about him before, so it's not just us. he's had to move from louisiana, to mississippi, to georgia, then here, because they didn't like his teaching. i mean.... wtf. our school is so fucking stupid. ((*sniffle* one time i told him it was my birthday and he said "i don't care, go sit down".... ))

 

Fact: quadratic equations are the best thing ever to happen in algebra. J/k... I wish I remembered how to graph those things. My alg. teacher looked like Yoda, so it was kind of difficult to pay attention.

 

What you need to do is find your zeroes. If you got it down to (3x-7) and (3x-5) your in the right direction.

3x-7=0
3x=7
x=7/3
so your x is 7/3 and your y is 0 (7/3,0)

3x-5=0
3x=5
x=5/3
Likewise, you point is (5/3,0)

now you have your intercepts.

eh, just noticed that its been answered allready =P

Isn't math fun?

 

Its alright guys, I totally understand!!! Thanks for all of your help, for real! I don't know why I didn't get it before, its so simple and once you get it its actually enjoyable (in some really nerdy way).

 

Hahaha... I always get like that when I understand math.

 

just wait till you get into yr 12 and you get into the nitty gritty calculus stuff. without sounding like a wanker, its really fun! if you need any help i just finished my high school certificate, so give us a whirl. man its a shit trying to type out equations on the computer though!