Undefined
Since a vertical line goes straight up and down, its slope is undefined. Also, the x value of every point on a vertical line is the same. Therefore, whatever the x value is, is also the value of 'b'.x=bslope intercept form: y=mx+b
Veritcal asymptotes are where the denominator of a fraction becomes 0 and the value of f(x) becomes undefined. Set the denominator to 0 and solve.Horizontal asymptotes are values of f(x) when x→∞ and x→-∞Two examples:f(x)=3x² / (x²+1)Vertical:Set (x²+1)=0 and solve for x.x²=-1 has no answers, so there is no vertical asymptote.Horizontal:Divide all terms by the highest power of x to eliminate unimportant valuesdividing by x², you get 3 / (1 + (1/x²))As x→∞ then 1/x² vanishes, leaving 3/1=3, so there is an asymptote at y=3f(x)=(x-3) / (x²+3x)Vertical:Set (x²+3x)=0 and solve. x={0,-3} so there are vertical asymptotes at 0 and -3Horizontal:Divide out by x²(x/x² - 3/x²) / (x²/x² + 3x/x²) as x→∞The terms in the numerator all vanish, making the answer 0, so there is a horizontal asymptote at 0.Enjoy.■A vertical asymptote also exists for the value of x (assuming a function of x) where the function becomes undefined. For instance:f(x) = ln (x). The logarithmic function is not defined for x
A function y = f(x) has a vertical asymptote at x = c if,f(x) is continuous for values of x just above c and the value of f(x) becomes infinitely large or infinitely negative (but not oscillating between them) as x approaches c from above. The function could behave similarly as x approaches c from below.In such a case f(c) is a singularity: the function is not defined at that point.
An oblique asymptote is another way of saying "slant asymptote."When the degree of the numerator is one greater than the denominator, an equation has a slant asymptote. You divide the numerator by the denominator, and get a value. Sometimes, the division pops out a remainder, but ignore that, and take the answer minus the remainder. Make your "adapted answer" equal to yand that is your asymptote equation. To graph the equation, plug values.
Undefined
Every function has a vertical asymptote at every values that don't belong to the domain of the function. After you find those values you have to study the value of the limit in that point and if the result is infinite, then you have an vertical asymptote in that value
It will have the same asymptote. One can derive a vertical asymptote from the denominator of a function. There is an asymptote at a value of x where the denominator equals 0. Therefore the 3 would go in the numerator when distributed and would have no effect as to where the vertical asymptote lies. So that would be true.
No. The fact that it is an asymptote implies that the value is never attained. The graph can me made to go as close as you like to the asymptote but it can ever ever take the asymptotic value.
An undefined slope will always be a vertical line through any point of choice where the value of x = 0; while the y value changes indefinitely, the value of x will always equal zero, and therefore is undefined.
No. A horizontal line has a slope of 0. A vertical line has an undefined slope. Let's say you have a vertical line with the points (3,-2) and (3,4). The slope is (y1-y2)/(x1-x2) = (-2-4)/(3-3) = -6/0 = undefined. Notice that x is the same for both points. A line in which all points have the same value for x is a vertical line and as such has an undefined slope.
Since a vertical line goes straight up and down, its slope is undefined. Also, the x value of every point on a vertical line is the same. Therefore, whatever the x value is, is also the value of 'b'.x=bslope intercept form: y=mx+b
Good question. They have infinite slopes. Incidentally, vertical lines are not functions in the formal sense. One x value is mapped on to an infinite number of y values. A function requires that each x maps on to only one y value.
Veritcal asymptotes are where the denominator of a fraction becomes 0 and the value of f(x) becomes undefined. Set the denominator to 0 and solve.Horizontal asymptotes are values of f(x) when x→∞ and x→-∞Two examples:f(x)=3x² / (x²+1)Vertical:Set (x²+1)=0 and solve for x.x²=-1 has no answers, so there is no vertical asymptote.Horizontal:Divide all terms by the highest power of x to eliminate unimportant valuesdividing by x², you get 3 / (1 + (1/x²))As x→∞ then 1/x² vanishes, leaving 3/1=3, so there is an asymptote at y=3f(x)=(x-3) / (x²+3x)Vertical:Set (x²+3x)=0 and solve. x={0,-3} so there are vertical asymptotes at 0 and -3Horizontal:Divide out by x²(x/x² - 3/x²) / (x²/x² + 3x/x²) as x→∞The terms in the numerator all vanish, making the answer 0, so there is a horizontal asymptote at 0.Enjoy.■A vertical asymptote also exists for the value of x (assuming a function of x) where the function becomes undefined. For instance:f(x) = ln (x). The logarithmic function is not defined for x
Because in a vertical line the slope is undefined, there is no "y" answer or "b" value and the line is in the form of x = some number such as x = 3 which is a vertical line.
In terms of the slope intercept form of the line, it is undefined. A vertical line is not a function since a single value of x is mapped onto infinitely many y values.
Yes, the asymptote is x = 0. In order for logarithmic equation to have an asymptote, the value inside log must be 0. Then, 5x = 0 → x = 0.