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Question 630349: Find the intercepts and asymptotes, and then sketch a graph of the rational function.
2x−4/
x2−7x+10
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! Please- Provide the entire problem. You left out the y or f(x) and the equals sign.
- Put multiple-term numerators and denominators in parentheses.
- Use "^" (shift+6) to indicate exponents.
Badly posted problems are less likely to get a response.
Look for "holes" first. (You don't mention holes but this graph has one and it is important to find holes first. It affects how we find x-intercepts and vertical asymptotes.) A "hole", if any, occurs when the function's fraction can be reduced by a factor that could be zero. If there is such a factor, then the x value that makes this factor zero will he a hole in the graph. To look for holes you factor the numerator and denominator and see if they have a common factor what could be zero:
We we can see, x-2 is a common factor. And it would be zero if x was 2. So we do have a hole and it will occur when x=2.
y-intercept (where the graph intercepts the y-axis) has an x coordinate of zero. To find it just make the x zero and solve for y:
which simplifies to -4/10 or -2/5. So the y-intercept is (0, -2/5)
x-intercept (where the graph intercepts the x-axis) has a y coordinate of zero. To find it just make the y zero and solve for x:
A fraction is zero only when its numerator is zero. So
2x-4 = 0
Solving this we get x = 2. But this is our hole. So there is no x-intercepts on this graph!
The vertical asymptotes, if any, occur for x values that make a denominator zero (other than any holes). To find them, set the denominator equal to zero and solve for x:
This is a quadratic equation. So we'll factor it (or use the Quadratic Formula). This factors easily:
(x-5)(x-2) = 0
From the Zero Product Product Property we know that one of the factors must be zero:
x-5 = 0 or x-2 = 0
Solving these we get:
x = 5 or x = 2
The 2 is our hole. So the only vertical asymptote is x = 5.
Other asymptotes, if any, occur for large positive or negative values of x's (way off to the right or left on the graph). To find them you have to analyze the function for high values. One way to do this is to divide the numerator and denominator by the highest power of x in the function. For this equation, that would be  :
which simplifies to:
This version of the equation helps us figure out what happens to y for large x's. When x has very large values, all those little fractions will have very large denominators. And fractions with very large denominators are very small numbers. In fact, the larger x gets the closer the fraction as a whole gets to zero! So if we replace replace all those little fractions with an x in the denominator with zeros we can see the value that y approaches for large x's:
So y = 0 (the x-axis) is the horizontal asymptote.
The y-intercept, two asymptotes and a hole are not much to go on. To supplement this:- If you know some Calculus, you could use the first and second derivatives to find concavity, relative maximums and minimums and points of inflection, if any.
- Find some extra points. Build a table of values by picking values for x and then using the equation to find its y value. Each of these will be another point on the graph. You can pick any numbers for the x values except those of the vertical asymptotes: 2 and 6. I would suggest numbers near the asymptotes and the hole, 1, 3, 5, 7 and large values (positive and negative). Whatever x values you pick plot the points and keep going until you can "see" how the graph goes. Then draw a smooth curve that connects the dots.
As you draw your graph remember the hole! As you draw your curve, put an small open circle (a small "oh") where x=2 on the curve. You should get a graph that looks something like this:
Notes on the graph:- Algebra.com's graphing software does not draw holes well. I adjusted the scale of the graph so you might be able to see it. But all you see is a little bump, not a small open circle at x = 2.
- The asymptotes should be drawn with dotted lines. But I cannot figure out how to get algebra.com's graphing software to draw a dotted line.
- Algebra.com's graphing software is not perfect. The graph may look like it intercepts the vertical asymptote. It does not. Graphs never cross the vertical asymptotes!
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