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klio [65]
3 years ago
8

The graph of a quadratic function with vertex (1,-3) is shown In the figure below . Find the domain and the range

Mathematics
1 answer:
forsale [732]3 years ago
6 0

Answer:

We can not see the graph, but that does not matter.

For a function y = f(x), the domain is the set of the possible values of x that we can input in the function, and the range is the set of the possible outputs.

First, for the domain, we start assuming that the domain is the set of all real numbers.

Then we look at our function, there is a point that causes a problem?

(a problem can be a zero in a denominator, for example)

Well, we have a quadratic equation, then we can not have any "x" in a denominator, then there are not problems, which means that the domain is the set of all real numbers.

For the range we first need to find the minimum (or maximum value) of f(x).

Here we know that the vertex is (1, - 3)

If the arms of the graph open upwards, then the minimum is at the vertex, which means that the minimum value of y is y = -3

And the arms go upwards infinitely, so there is no upper limit.

Then the range is the set of all numbers such that:

-3 ≤ y

Now, if the arms of the graph open downwards, then the maximum is at the vertex, which means that the maximum of our function is at y = -3

And with similar thinking than in the above case, we can find that the range is:

R:  y ≤ -3.

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daser333 [38]

Answer:

a) Number of projects in the first year = 90

b) Earnings in the twelfth year = $116500

Total money earned in 12 years = $969000

Step-by-step explanation:

Given that:

Number of projects done in fourth year = 129

Number of projects done in tenth year = 207

There is a fixed increase every year.

a) To find:

Number of projects done in the first year.

This problem is nothing but a case of arithmetic progression.

Let the first term i.e. number of projects done in first year = a

Given that:

a_4=129\\a_{10}=207

Formula for n^{th} term of an Arithmetic Progression is given as:

a_n=a+(n-1)d

Where d will represent the number of projects increased every year.

and n is the year number.

a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)

Subtracting (2) from (1):

78 = 6d\\\Rightarrow d =13

By equation (1):

129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90

<em>Number of projects in the first year = 90</em>

<em></em>

<em>b) </em>

Number of projects in the twelfth year =

a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233

Each project pays $500

Earnings in the twelfth year = 233 \times 500 = $116500

Sum of an AP is given as:

S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938

It gives us the total number of projects done in 12 years = 1938

Total money earned in 12 years = 500 \times 1938 = $969000

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