6. find any points of discontinuity for the rational function. y = x-8/x^2 +6x-7 x = 1, x = 7 x = 8 x = 1, x = –7 x = –1, x =

Question

Identify any points of discontinuity for the rational function ( y = \frac{x - 8}{x^2 + 6x - 7} ). Consider the values ( x = 1 ), ( x = 7 ), ( x = 8 ), ( x = -7 ), and ( x = -1 ).

Describe the vertical asymptotes and holes for the graph of ( y = \frac{x - 5}{x^2 + 4x + 3} ). Consider the following options for asymptotes and holes:

Asymptotes: ( x = -3 ), ( x = -1 ); No holes.
Asymptote: ( x = -3 ); Hole: ( x = -5 ).
Asymptotes: ( x = -3 ), ( x = -1 ); Hole: ( x = -5 ).

Asymptote: ( x = -5 ); Hole: ( x = -3 ).

Find the horizontal asymptote of the graph of ( y = \frac{-4x^6 + 6x + 3}{8x^6 + 9x + 3} ). Consider the following possibilities:

( y = 1 )
( y = -\frac{1}{2} )
( y = 0 ) 
There is no horizontal asymptote.

Anonymous · Accepted Answer

6. Find any points of discontinuity for the rational function.y = x-8/x^2 +6x-7 x = 1, x = –77. Describe the vertical asymptote(s) and hole(s) for the graph of y = x-5/x^2 +4x +3asymptotes: x = –3, –1 and no holes. 8. Find the horizontal asymptote of the graph of y = -4x^6 +6x +3/8x^6 +9x+3There is no horizontal asymptote.

Anonymous · Answer

6) Points of discontinuity of a rational function are those points where the denominator is equal to zero. Therefore you need to solve the equation:x² + 6x - 7 = 0(x + 7)(x - 1) = 0x = -7 and x = -1 are points of discontinuity7) Vertical asymptotes are found at those points where the denominator is equal to zero, holes are found at those points where both numerator and denominator are equal to zero at the same time.Therefore you need to solve two equations:x - 8 = 0x = 8x² + 4x + 3 = 0(x + 3)(x + 1)x = -3 and x = -1Therefore, you have no holes and x = -3 and x = -1 are vertical asymptotes.8) In order to find a horizontal asymptote, you need to look at the degree of both numerator and denominator and they must be equal to each other.In your case, both degrees are 6, therefore you have a horizontal asymptote.In order to find which line is the horizontal asymptote, you need to divide the coefficients of the highest degree monomials: -4/8 = -1/2. Therefore your horizontal asymptotes is y = -1/2

Anonymous · Answer

x=-5, x=1 Step-by-step explanation:y=x-8/x^2+4x-5 When denominator becomes 0 in a rational function then there will be a break in the graph To find any points of discontinuity for the rational function, we set the denominator =0 and solve for xx^2 + 4x -5 =0Now we factor x^2 +4x -5product is -5  and sum = 45 * (-1) = -55 +(-1) = 4(x+5)(x-1) =0set each factor =0 and  solve for xx+5=0 , so x= -5 x-1=0 , so x= 1 x=-5, x=1 are the points of discontinuity for the rational function

Anonymous · Answer

Option C. x = –9, x = –7Step-by-step explanation:we know thatIn the rational function, the denominator cannot be equal to zerosoIn this problemwe have that the denominator is equal to equate to zerosoThe solutions is  x=-9 and x=-7soFor x=-7 or x=-9 the denominator is equal to zerothereforex=-9, x=-7 are points of discontinuity for the rational function given

Anonymous · Answer

The answer is x=-1, x=8

Anonymous · Answer

The function is not continuous at x= -1 and x= 8Step-by-step explanation:The given function is,Factoring the denominator,Now the function becomes,The function is not defined (it does not exist) for x= -1 and for x= 8, because the denominator is zero for those values of x.To be continuous, the function has to be defined. So at those points the function is not continuous.*At x= -1 and at x= 8 , the function has a vertical asymptote.

Mr. Lazaro Wuckert III · Answer

A point of discontinuity is any point that yields an undefined answer when plugged in (meaning there’s division by zero involved).By that, we can find the points of discontinuity by setting the denominator equal to 0.x^2 + 5x - 6 = 0(x + 6)(x - 1) = 0x = 1, -6Those are your points of discontinuity

Mitchell Legros I · Answer

Factorit simplifies to y=1(x-7)(x-3)they are undefined since they make the denomenator 0solve(x-7)(x-3)=0x-7=0x=7x-3=0x=3restricted values are x=3 and x=7but, the points x=3 and x=7 are holes since they were canceled outanswer is B

Anonymous · Answer

For discontinuity of the function:x² - 7 x - 8 ≠ 0x² - 8 x + x - 8 = 0x ( x - 8 ) + ( x - 8 ) ≠ 0( x - 8 ) ( x + 1 ) ≠ 0The points of discontinuity are: x = - 1 and x = 8.As for the Domain of the function:x ∈ ( - ∞, - 1 ) ∪ ( - 1 , 8 ) ∪ ( 8, +∞ ).

Anonymous · Answer

For  y = (x^2 -4)/((x +2)(x^2 -49))the numerator factors to (x -2)(x +2), so the factor of (x +2) will cancel with that in the denominator, leaving  y = (x -2)/(x^2 -49)There are points of discontinuity at the hole, x=-2, and at each of the vertical asymptotes, at x=-7, +7.The horizontal asymptote is y=0.

Graciela Roob DDS

6. find any points of discontinuity for the rational function. y = x-8/x^2 +6x-7 x = 1, x = 7 x = 8 x = 1, x = –7 x = –1, x =

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