Answer:
domain = ( 3, 11, 121, 34, 23)
range = ( 21, 34, 1, 23)
Answer: 24
A hyperbola is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point in the same plane to its distance from a fixed line is always constant, which is always greater than unity.
the fixed point is called the focus and the fixed line is directrix and the ratio is the eccentricity.
The general equation for the vertical hyperbola is
[ (y-k)^2 / a^2 ] – [ (x-h)^2 / b^2 ] = 1
The conjugate axis of the vertical hyperbola is y = k
Length of the conjugate axis = 2b
According to the question k = 2, h = -1, a = 4, b = 12
Length of the conjugate axis = 2b = 2 * 12 = 24
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Answer: has no equal sign
Answer:
Step-by-step explanation:
Vertical Asymptote: x=2Horizontal Asymptote: NoneEquation of the Slant/Oblique Asymptote: y=x 3+23 Explanation:Given:y=f(x)=x2−93x−6Step.1:To find the Vertical Asymptote:a. Factor where possibleb. Cancel common factors, if anyc. Set Denominator = 0We will start following the steps:Consider:y=f(x)=x2−93x−6We will factor where possible:y=f(x)=(x+3)(x−3)3x−6If there are any common factors in the numerator and the denominator, we can cancel them.But, we do not have any.Hence, we will move on.Next, we set the denominator to zero.(3x−6)=0Add 6 to both sides.(3x−6+6)=0+6(3x−6+6)=0+6⇒3x=6⇒x=63=2Hence, our Vertical Asymptote is at x=2Refer to the graph below:enter image source hereStep.2:To find the Horizontal Asymptote:Consider:y=f(x)=x2−93x−6Since the highest degree of the numerator is greater than the highest degree of the denominator,Horizontal Asymptote DOES NOT EXISTStep.3:To find the Slant/Oblique Asymptote:Consider:y=f(x)=x2−93x−6Since, the highest degree of the numerator is one more than the highest degree of the denominator, we do have a Slant/Oblique AsymptoteWe will now perform the Polynomial Long Division usingy=f(x)=x2−93x−6enter image source hereHence, the Result of our Long Polynomial Division isx3+23+(−53x−6)
