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lesantik [10]
3 years ago
11

Factorise fully 3x-9x^2

Mathematics
1 answer:
artcher [175]3 years ago
6 0

Answer:

3x(1 - 3x)

Step-by-step explanation:

We can factor the expression by using the common factor.

Pull out 3x from the expression.

3x - 9 {x}^{2}  \\ 3x(1 - 3x)

Common Factor is similar to dividing by the term we pull out. When we pull out the term, we divide the whole expression and keep the term out of the expression.

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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

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Evaluate the expression 3(x-1)2nd power +2x-7 for x=3
Yanka [14]

Answer:

11

3Step-by-step explanation:

Given

3(x - 1)² + 2x - 7 ← substitute x = 3 into the expression

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= 11

5 0
3 years ago
I need help pleaseeeeee​
MA_775_DIABLO [31]

Answer:

D

Step-by-step explanation:

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