How do you find holes in rational functions?
The term "hole" used here is another name for a removable discontinuity or removable singularity.A rational function [math]f(x)=\dfrac{p(x)}{q(x)}[/math] is the quotient of two polynomials. Of course, when the denominator [math]q(x)[/math] is 0, then [math]f(x)[/math] is not defined. That is to say, if [math]q(a)=0,[/math] the number [math]a[/math] is not in the domain of [math]f(x).[/math]However, if you find that [math](x-a)[/math] is a factor of both the numerator and denominator, and it occurs in the numerator with as great a multiplicity as it does in the denominator, then you can cancel it. The denominator will no longer have the value 0 at [math]a,[/math] and [math]f(x)[/math] will become defined at [math]a.[/math] You've filled in the hole.So, to find and fill in the holes, factor the numerator and denominator, and cancel common factors.Actually, you don’t have to resort to factoring. Instead, you can use Euclid’s algorithm to find the greatest common divisor (GCD) of the numerator and the denominator, then divide each by that GCD. The algorithm works for polynomials in the same way that it works for positive integers. Divide the polynomial with the lower degree into the one with the higher degree, then replace the higher degree one by the remainder.
Polynomial and Rational Functions homework help?
Let x be the length of one of the five parallel lengths of fence. Then (50000 - 5x) / 2 is the length of one of the two longer rectangular sides. Total area = x ((50000 - 5x) / 2) Simplify the area expression: 25000x - (5/2)x^2 Take the derivative and set it to zero: 25000 - 5x = 0 Solve for x: x = 5000 Substitute this value for x into the expression for the total area and evaluate: 5000 ((50000 - 5(5000)) / 2) = 62500000 square feet total Divide by four to determine the area of one of the four pastures: 62500000 / 4 = 15625000 square feet
Precal linear to linear help please!?
Oscar is hunting magnetic fields with his gauss meter, a device for mea- suring the strength and polarity of magnetic fields. The reading on the meter will increase as Oscar gets closer to a magnet. Oscar is in a long hallway at the end of which is a room containing an extremely strong magnet. When he is far down the hallway from the room, the meter reads a level of 0.2. He then walks down the hallway and enters the room. When he has gone 6 feet into the room, the meter reads 2.3. Eight feet into the room, the meter reads 4.4. (a) Give a linear-to-linear rational model re- lating the meter reading y to how many feet x Oscar has gone into the room. (b) How far must he go for the meter to reach 10? 100? (c) Considering your function from part (a) and the results of part (b), how far into the room do you think the magnet is?
PreCalculus polynomial and rational function help?
Let f(x) = (1/50,000) x^3 + (1/2) x a) If x is very small (less than 1), then x^3 gets smaller and smaller. For example if x = 0.1, x^3 = 0.001 So the term (1/2)x will be more important as (1/50000)x^3 will approach the value of Zero b) This asks to calculate the values of y = (1/50,000) x^3 + (1/2) x from x = - 10 to 10 at regular intervals; say at x = -10, x= -8, x = -6 etc. Then you plot only the points where the value of y lies between -10 and 10. c) If (1/50,000) x^3 = (1/2) x x^3 /50000 = x /2 Multiply both sides by 2/x: x^2 / 25000 = 1 x^2 = 25000 Value of x will be = √25000 = 158.1
How can I master slant asymptotes and trig functions in precalculus?
It’s hard to offer much in the way of specifics, given the relatively broad range of material that you’re having trouble with. Polynomial division (long and synthetic) is relatively narrow, and mastering those will make short work of slant asymptotes. However, “trig functions”, in the curricula that I taught, spans a quarter to a third of the entire course. (You might need to be a little more specific.)In general, though, if your teacher isn’t doing a good job of helping you understand those particular topics, see if another pre-Calc teacher at your school can make some time for you, or look into peer tutoring services, if your school offers those. Outside of school, there are numerous websites that offer online tutoring services, generally for a small fee, and you can probably find local tutors who could help out in person, after school or on weekends, also for a fee.If you have specific questions or problems you’d like help with, feel free to comment here, and I’ll see what I can do. :-)
Pre-Calculus Polynomial function f(x)= 4x^5-8x^4-5x^3+10x^2+x-2?
f(x)= 4x^5-8x^4-5x^3+10x^2+x-2 = 4x^4(x-1)-4x^3(x-1)-9x^2(x-1)+x(x-1) +2(x-1)= (x-1)(4x^4-4x^3-9x^2+x +2)= (x-1)(4x^3(x+1) -8x^2(x+1) -x(x+1) +2(x +1))= (x-1)(x+1)(4x^3-8x^2-x+2)= (x-1)(x+1)(4x^2(x-2)-(x-2))= (x-1)(x+1)(x-2)(4x^2-1) (x-1)(x+1)(x-2)(2x-1)(2x+1) zeroes -1, -1/2, 1/2, 1, 2
Find the linear-to-linear function whose graph has y = 6 as a horizontal asymptote and passes through (0, 12) and (5, 7).?
Find the linear-to-linear function whose graph has y = 6 as a horizontal asymptote and passes through (0, 12) and (5, 7).? A linear-to-linear function is a rational function in the form f(x) = (ax+b)/(x+c) =======================================... In a rational function, if the degree of the numerator is the same as the degree of the denominator then it has a horizontal asymptote of the form y = k where k is the ratio of the leading co-efficients in the numerator and the denominator; k = 6, k = a/1, a = 6. Now we have: f(x) = (6x+b)/(x+c) and the points (0, 12) and (5, 7). Plug in the points to the function to form two equations in two unknowns: 12 = b/c and 7 = (30+b)/(5+c) From the first, b = 12c, substitute into the second: 7 = (30+12c)/(5+c) Solve for c: 35 + 7c = 30 + 12c 5 = 5c, c = 1 b = 12c = 12*1 = 12 f(x) = (6x+12)/(x+1)
Is learning algebra 2 needed to learn calculus?
I would assert that the concepts of algebra II are not necessarily central to the concepts of calculus. You can understand derivatives and integrals without knowing, say, the quadratic formula.However, the calculations of AlgII (and most of Pre-Calc) are absolutely vital to the calculations of Calculus. You learn all sorts of basic information about a variety of functions (rational functions, ln, sin/cos…), rules (log/exponential rules, trig identites), and techniques (partial fraction decomposition) which you have to use all the time in Calc to take derivatives and integrals.Therefore, the class of AlgII is necessary for the class of Calculus, because to be able to actually do calculus (rather than just understand calculus), you need a variety of tricks from ordinary algebra.So: definitely take Algebra II in advance of Calculus, unless you know everything from it (ask a teacher at your school; exact curricula can vary subtly). However, fell free to start reading about calculus on your own (or asking your teachers about them, etc.), because you can probably understand the underlying concepts, even if specific computations are beyond your grasp.