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olga_2 [115]
3 years ago
10

What is the graph of the function f(x) = the quantity of x squared plus 5x plus 6, all over x plus 3?

Mathematics
2 answers:
Andrei [34K]3 years ago
6 0
Positive sloping linear graph with hole at negative 3, negative 1 intersecting the x axis at negative two and the y axis at two
velikii [3]3 years ago
5 0

Answer:

The graph of given function is shown below.

Step-by-step explanation:

The given function is

f(x)=\frac{x^2+5x+6}{x+3}

Factorize the numerator.

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

f(x)=\frac{x(x+3)+2(x+3)}{x+3}

f(x)=\frac{(x+2)(x+3)}{x+3}

Cancel out the common factor.

f(x)=x+2

It is a straight line because it is a one degree polynomial.

At x=0,

f(0)=0+2=2

The y-intercept of the function is (0,2).

Put f(x)=0

0=x+2

x=-2

The x-intercept of the function is (-2,0).

Equate the cancel factor equal to 0, to find the hole of the function.

x+3=0\Rightarrow x=-3

The function has hole at x=-3.

The graph of given function is shown below.

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Answer:

The next number of the series 0, 1/3, 1/2, 3/5, and 2/3 is 5/7

Step-by-step explanation:

The given numbers are;

0, 1/3, 1/2, 3/5, and 2/3

The number sequence is formed adding \dfrac{1}{\left (\dfrac{n^2 + n}{2} \right ) } to each (n - 1)th term to get the nth term number in the sequence, with the first term equal to 0, as follows;

For the 2nd term, the (n - 1)th term is 0, and n = 2, gives;

The

0 +\dfrac{1}{\left (\dfrac{2^2 + 2}{2} \right ) } = 0 + \dfrac{1}{3} = \dfrac{1}{3}

For the 3rd term, the (n - 1)th term is 1/3, and n = 3, gives;

\dfrac{1}{3} +\dfrac{1}{\left (\dfrac{3^2 + 3}{2} \right ) } = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{2}

For the 4th term, the (n - 1)th term is 1/2, and n = 4, gives;

\dfrac{1}{2} +\dfrac{1}{\left (\dfrac{4^2 + 4}{2} \right ) } = \dfrac{1}{2} + \dfrac{1}{10} = \dfrac{3}{5}

For the 5th term, the (n - 1)th term is 3/5, and n = 5, gives;

\dfrac{3}{5} +\dfrac{1}{\left (\dfrac{5^2 + 5}{2} \right ) } = \dfrac{3}{5} + \dfrac{1}{15} = \dfrac{2}{3}

For the next or 6th term, the (n - 1)th term is 2/3, and n = 6, gives;

\dfrac{2}{3} +\dfrac{1}{\left (\dfrac{6^2 + 6}{2} \right ) } = \dfrac{2}{3} + \dfrac{1}{21} =  \dfrac{15}{21} = \dfrac{5}{7}

The next number of the series 0, 1/3, 1/2, 3/5, and 2/3 = 5/7.

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Step-by-step explanation:

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