How can i draw the graph of f(x)=2x^2-3x+2/x^2+1 please help me
hard to tell if you mean
f(x)=2x^2-3x+2/x^2+1
or
f(x)=2x^2-3x+2/(x^2+1)
or
f(x)=(2x^2-3x+2)/x^2+1
or
f(x)=(2x^2-3x+2)/(x^2+1)
in any case, make a table of values using reasonable values of x to find the corresponding y values.
Plot your points, join them with a smooth curve
You can test your graph here
http://rechneronline.de/function-graphs/
Make sure you only enter the right side of your equation.
What is the horizontal asymptote? I think it doesn't have vertical and oblique asymptote. Please please help me please
Did you not use the webpage I suggested?
You would be able to answer your last questions by looking at the graph.
You also did not answer my questions as to what the function really is
I will assume the last version I posted.
the horizontal asymptote is y = 2
The webpage gragh and the graph that our teacher draw is not the same what shall i do?
To draw the graph of the function f(x) = (2x^2 - 3x + 2) / (x^2 + 1), we can follow these steps:
1. Determine the domain restrictions: In this case, there are no restrictions because the denominator, x^2 + 1, is always positive.
2. Find the asymptotes: To find the vertical asymptotes, set the denominator equal to zero and solve for x. Since x^2 + 1 does not equal zero for any real value of x, there are no vertical asymptotes in this case. However, there is a horizontal asymptote, which we can find using the degrees of the numerator and denominator. Since the degree of the numerator (2x^2 - 3x + 2) is equal to the degree of the denominator (x^2 + 1), the horizontal asymptote will be y = the ratio of the leading coefficients. In this case, y = 2/1 = 2.
3. Find the x-intercepts: To find the x-intercepts, set the numerator equal to zero and solve for x. So, 2x^2 - 3x + 2 = 0. We can factor this quadratic equation as (2x - 1)(x - 2) = 0, which gives us x = 1/2 and x = 2 as the x-intercepts.
4. Find the y-intercept: To find the y-intercept, substitute x = 0 into the function. So, f(0) = (2(0)^2 - 3(0) + 2) / ((0)^2 + 1) = 2.
Now, we have all the relevant points to plot the graph:
- Vertical asymptotes: None
- Horizontal asymptote: y = 2
- X-intercepts: (1/2, 0) and (2, 0)
- Y-intercept: (0, 2)
Using this information, plot these points on a coordinate plane, and then draw a smooth curve that approaches the asymptote and passes through the points. Note that since the equation is rational, the graph may have vertical gaps or holes, but that won't affect the overall shape of the graph.