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ArbitrLikvidat [17]
2 years ago
11

Any clue? Find the equation of the line shown.

Mathematics
1 answer:
icang [17]2 years ago
4 0

Answer:

Y= -X + 9 Can be found by trials

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Select the correct answer.
larisa [96]

Answer:

B

Step-by-step explanation:

In this graph we can see a "Parabola", this is the curve for a second degree polynomial function, and based on "Fundamental theorem of algebra" we can know that this polynomial has 2 roots (they can be real or imaginary).

In this graph, the curve doesn't touch the X axis, so we know that this function has not real root. So both roots are complex

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=F%28x%29%20%3D%20%5Cfrac%7Bx%5E%7B3-8%7D%20%7D%7Bx%5E%7B2%7D%20-6x%2B8%7D" id="TexFormula1" ti
Firlakuza [10]

Answer:

  • Domain: All the real values except x = 2 and x = 4: R - {2, 4}
  • Holes: x = 2
  • VA, vertical asymptores: x = 4
  • HA: horizontal asymptotes: there are not horizontal asymptotes
  • OA: oblique asymptotes: x + 6 [note that OH does not stand for any known feature, and so it is understood that it was intended to write OA]
  • Roots: x = 2
  • Y-intercept: -1

Step-by-step explanation:

1. <u>Given</u>:

f(x)=\frac{x^3-8}{x^2-6x+8}

  • Note that the number 8 in the numerator is not part of the power.
  • Type of function: rational function

2. <u>Domain</u>: is the set of x-values for which the function is defined.

The given function is defined for all x except those for which the denominator equals 0.

  • Denominator:  x² -6x + 8 = 0
  • Solve for x:

        Factor. (x - 4 )(x - 2) = 0

        Zero product property: (x - 4) = 0 or (x - 2) = 0

        x - 4 = 0 ⇒ x = 4

        x - 2 = 0 ⇒ x = 2

  • Domain:

        All the real values except x = 2 and x = 4: x ∈ R / x ≠ 2 and x ≠ 4.

3. <u>Holes</u>:

The holes on the graph of a rational function are at those x-values for which both the numerator and denominator are zero.

  • Find the values for which the numerator is zero:

        Numerator: x³ - 8 = 0

        Factor using difference of cubes property:

                   a³ - b³ = (a - b)(a² + ab + b²)

                   x³ - 8 = (x - 2)(x² + 2x + 4) = 0

        Zero product property:  (x - 2)(x² + 2x + 4) = 0

                    x - 2 = 0 ⇒ x = 2                    

                    x² + 2x + 4 = 0 (this has not real solution)

  • The values for which the denominator is zero were determined above: x = 2 and x = 4.

  • Conclusion: for x = 2 both numerator and denominator equal 0, so this is a hole.

4. <u>VA: Vertical asymptotes</u>.

The vertical asymptotes on the graph of a rational function are the vertical lines for which only the denominator (and not the numerator) equals zero.

  • In the previous part it was determined that happens when x = 4.

5. <u>HA: Horizontal asymptotes</u>.

In rational functions, if the numerator is a higher degree polynomial than the denominator, there is no horizontal asymptote.

6. <u>OA: oblique asymptotes</u>

  • Find the quotient and the remainder.

                       x + 6

                  _______________

x² - 6x + 8 )   x³ + 0x² + 0x - 8

                  - x³ + 6x² - 8x

                   ___________

                          6 x² -   8x -  8

                        - 6x² + 36x - 48

                        _____________

                                    28x  - 56

Result: (x + 6) + (28x - 56) / (x² - 6x + 8)

  • Find limit x → ∞

\lim_{x \to \infty}(x + 6) + \frac{28x-56}{x^2-6x+8}=x+6

<u>7. Roots</u>:

Roots are the values for which f(x) = 0.

That happens when the numerator equals 0, and the denominator is not 0.

As determined earlier: x³ - 8 = 0 ⇒ x = 2.

8. <u>Y-Intercept</u>

The y-intercepts of any function are the y-values when x = 0

  • Substitute x = 0 into the function:

         f(x)=\frac{x^3-8}{x^2-6x+8}=\frac{0^3-8}{0^2-6(0)+8}}=\frac{-8}{8} =-1

3 0
3 years ago
Which of the following quartic functions has x=-1 and x=-2 as its only two real zeros
Brums [2.3K]
Quartic is 4th degree
the factors of an equation with roots r1,r2 is
(x-r1)(x-r2)
4th degree
it could be
(x-r1)¹(x-r2)³ or
(x-r1)²(x-r2)² or
(x-r1)³(x-r2)¹


roots or zeroes at x=-1 and x=-2
(x-(-1)) and (x-(-2))
(x+1) and (x+2)

the function could be factored into
(x+1)¹(x+2)³ or
(x+1)²(x+2)² or
(x+1)³(x+2)¹

expanded would be
x⁴+7x³+18x²+20x+9 or
x⁴+6x³+13x²+12x+4 or
x⁴+5x³+9x²+7x+2
one of those is the answer

4 0
3 years ago
Pleaseeee helppp answer correctly !!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!!!!!
AnnZ [28]
2y+4=3(y-1)
2y+4=3y-3
4+3=3y-2y
Y=7
X=2
4 0
3 years ago
Read 2 more answers
Refer to the figure below to complete the following problem.
IRINA_888 [86]

Answer:

38

Step-by-step explanation:

5x-7=3x+ll

5x=3x+18

2x/2=18/2

x    =9

5 x 9 -7 = 36

3 x 9 + 11=36

m D =36

3 0
3 years ago
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