Find the vertical and horizontal asymptotes of the rational function f(x)=(2x^2-5)/(x^2-9)
Vertical asymptotes: To find the vertical asymptotes, we equate the denominator to 0 and solve for x. x^2-9=0 We factor it by using the difference of square formula a^2-b^2=(a+b)(a-b) So, x^2-3^2=0 (x+3)(x-3)=0 Equate each term to 0 and solve for x. x+3=0.........so, x=-3 x-3=0..........so, x=3 The vertical asymptotes are x=-3 and x=3. Horizontal asymptotes: To find the horizontal asymptotes, we look at the degree(exponent of the variable) of the numerator and the denominator. If the degree of the numerator and the denominator are equal, then we just divide the co-efficients of the highest degree term to get the horizontal asymptote. Degree of the numerator 2x^2-5 is 2. Degree of the denominator x^2-9 is 2. As the degree of the numerator and denominator is same, we divide the co-efficient of the highest degree term. So, y=2/1=2 y=2 is the horizontal asymptote.
Find the amplitude, period and phase shift of the function f(x)=2sin(3(x-pi/3))
First we start by comparing given function with the general sine function f(x)=Asin(B(x-C))+D where A=amplitude Period=(2pi)/B C is the phase shift and D is the vertical shift Comparing the given function f(x)=2sin(3(x-pi/3)) with the general sine function, we get Amplitude A=2 Period=(2pi)/B=(2pi)/3 Phase shift C=pi/3
The admission fee at a fair is $3 for children and $5 for adults. On a certain day, 2200 people enter the fair. The fair made $8400 from admission tickets that day. How many children and adults bought a ticket?
First we start by defining the variables for the number of tickets bought by the children and adults. Let number of tickets bought by the children be "x" and number of tickets bought by the adults be "y". Given that, 2200 people attended the fair. So, x+y=2200 Subtract x on both sides to get y by itself y=2000-x...........equation(1) The admission fee is $3 for children, $5 for adults and the fair made $8400 from the tickets. So, 3x+5y=8400......equation(2) Now substitute y=2000-x in the second equation 3x+5(2200-x)=8400 Distribute 5 to 2200-x term. 3x+10000-5x=8400 Combine like terms (3x-5x)+11,000=8400 -2x+11000=8400 Subtract 11,000 on both sides -2x=8400-11000 -2x=-2600 Divide by -2 on both sides x=1300 Now substitute the value of x in the first equation y=2200-1300 y=900 So, 1300 children and 900 adults attended the fair.