I know how to graph a function or a relation but what do you mean by graphing an equation?
In this equation, "1/(2x+1) - 1 = 4/(2x^2-5x-3)", how can I go about graphing it?
What I mean is, I understand how to factor it out and get the x-intercepts and vertical asymptotes but what I don't understand is how you know where the graph goes.
I ended up getting vertical asymptotes of -1/2, and 3, and x-intercepts of 2, and 1.
My answers book shows that is correct, but also shows a graph like this : i.imgur.com/qfzGcWZ.png
My question is how do they know that on the left it does not cross the horizontal asymptote and goes downward, how do they know in the middle it comes from the top and goes back up rather than coming from the bottom and going back down, and likewise for the right side? I have no idea and my book doesn't explain it at all.
Thank you for any help, please assume I'm not good at math and explain it as simply as possible if you can.
I mean the expression I wrote there, the function. like I said above, I'm not good at math so I will use incorrect terminology by mistake from time to time.
was it really impossible to understand what I meant?
$ let f(x) = \frac{4}{2x^2 - 5x - 3}
and
g(x) = \frac{1}{2x + 1}-1 $
To graph f(x):
1- Find where f(x) does not exist:
$ 2x^2 -5x - 3 = 0$
$(2x+1)(x-3) = 0$
So at x= -0.5 and x= 3 are our vertical asymptotes.
2- Are there any horizontal asymptotes? Yes at y=0.
3- How does our function approach the horizontal asymptote?
$f(1000) = 0.00021 $
$f(-1000) = 0.000001995 $
So as x approaches positive/negative infinity, f(x) approaches y=0 from above.
4- How does f(x) behave around the vertical asymptote?
At x = -0.5 : $\lim_{x \to -0.5^-} f(x) = 57142$ and $\lim_{x \to 0.5^+} f(x) = -0.8 $
At x = 3: $\lim_{x \to 3^-} f(x) = -571.6 $ and $\lim_{x \to 3^+} f(x) = 5714 $
So as: $ x \to 0.5 $ from the left, $y \to \infty $ and $ x \to 0.5 $from the right $y \to -\infty $
So as: $ x \to 3 $ from the left, $y \to -\infty $ and $ x \to 3 $from the right $y \to \infty $
Mhmmmm now I think about it have you done calculus? If yes, find the first derivative and then evaluate it where $f'(x) = 0 $ or does not exist then check whether $f'(x)$ increases or decreases between those intervals which will tell you where $f(x)$ is increasing or decreasing.
Then repeat all of this for $g(x)$ and then after you have sketched both functions, evaluate the co-ordinates at which they intersect.
ORRRR just use this online graphing calculatorSpoiler:
4/(2x^2 -5x -3) = 1/(2x+1) -1 - Wolfram|Alpha
Yes, it is! A graph requires at least two parameters, usually x and y. Your equation is satisfied for x= 1 and 2 and only those.
There is no second parameter or variable. If you meant "plot these values on a number line", which would be a possible, though unusual, interpretation of "graph" you would just mark the points 1 and 2 on the line.
What Sakopure6 did was interpret your equation as two separate functions and tell you how to graph each one. But it is impossible to tell, from what you said, what you really meant.
I see, my mistake. I said that because the question in my book is "Solve for x in each of the rational equations, solve algebraically for the exact values of their roots," etc, and then I have to graph it. Does that mean sakonpure's response isn't correct for what my problem is?