
An engraver charges a \[\$ 10\] fee plus \[\$ 6\] for each line of engraving. Write a function to describe the situation. Find a reasonable domain and range for the function for up to 8 lines.
Answer
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Hint: To write the function of the given situation, first find the independent variable and dependent variable and then find how dependent variable is dependent on the independent variable to get the equation/function. To find the domain, put the value given as the end point and start it from one, because there cannot be zero or negative number of lines, after getting domain, put start and end point of domain in the function to get the range.
Complete step by step answer:
In the given question, the number of lines of engraving is the independent variable $\left( {{\text{say}}\;x} \right),$ because the engraver charges on the basis of it. So Engraver’s charge will be the dependent variable$\left( {{\text{say}}\;y} \right),$
Now, it is given that he charges \[\$ 6\] for each line of engraving,
That is for one line, $y = 6$
Therefore, for $x$ number of lines, $y = 6 \times x$
But in the question, it is also given that he charges a base price of \[\$ 10,\] which means this is constant whether the number of lines are $1\;{\text{or}}\;1000$
We can write this as
$
y = 6 \times x + 10 \\
y = 6x + 10 \\
$
Since number of lines cannot be zero or negative or fraction, so domain $ = \left\{ {1,\;2,\;3,\;4,\;5,\;6,\;7,\;8} \right\}$
Putting each value of the domain in the function,
\[
{y_1} = 6 \times 1 + 10 \\
{y_1} = 6 + 10 \\
{y_1} = 16 \\
{y_2} = 6 \times 2 + 10 \\
{y_2} = 12 + 10 \\
{y_2} = 22 \\
{y_3} = 6 \times 3 + 10 \\
{y_3} = 18 + 10 \\
{y_3} = 28 \\
{y_4} = 6 \times 4 + 10 \\
{y_4} = 24 + 10 \\
{y_4} = 34 \\
{y_5} = 6 \times 5 + 10 \\
{y_5} = 30 + 10 \\
{y_5} = 40 \\
{y_6} = 6 \times 6 + 10 \\
{y_6} = 36 + 10 \\
{y_6} = 46 \\
{y_7} = 6 \times 7 + 10 \\
{y_7} = 42 + 10 \\
{y_7} = 52 \\
{y_8} = 6 \times 8 + 10 \\
{y_8} = 48 + 10 \\
{y_8} = 58 \\
\]
Therefore range $ = \left\{ {16,\;22,\;28,\;34,\;40,\;46,\;52,\;58} \right\}$
Note: In solving this type of question, make sure the constant part is also written in the function, because it happens most of the times that students forget the constant part and just write the function or the equation with dependent and independent variables.
Complete step by step answer:
In the given question, the number of lines of engraving is the independent variable $\left( {{\text{say}}\;x} \right),$ because the engraver charges on the basis of it. So Engraver’s charge will be the dependent variable$\left( {{\text{say}}\;y} \right),$
Now, it is given that he charges \[\$ 6\] for each line of engraving,
That is for one line, $y = 6$
Therefore, for $x$ number of lines, $y = 6 \times x$
But in the question, it is also given that he charges a base price of \[\$ 10,\] which means this is constant whether the number of lines are $1\;{\text{or}}\;1000$
We can write this as
$
y = 6 \times x + 10 \\
y = 6x + 10 \\
$
Since number of lines cannot be zero or negative or fraction, so domain $ = \left\{ {1,\;2,\;3,\;4,\;5,\;6,\;7,\;8} \right\}$
Putting each value of the domain in the function,
\[
{y_1} = 6 \times 1 + 10 \\
{y_1} = 6 + 10 \\
{y_1} = 16 \\
{y_2} = 6 \times 2 + 10 \\
{y_2} = 12 + 10 \\
{y_2} = 22 \\
{y_3} = 6 \times 3 + 10 \\
{y_3} = 18 + 10 \\
{y_3} = 28 \\
{y_4} = 6 \times 4 + 10 \\
{y_4} = 24 + 10 \\
{y_4} = 34 \\
{y_5} = 6 \times 5 + 10 \\
{y_5} = 30 + 10 \\
{y_5} = 40 \\
{y_6} = 6 \times 6 + 10 \\
{y_6} = 36 + 10 \\
{y_6} = 46 \\
{y_7} = 6 \times 7 + 10 \\
{y_7} = 42 + 10 \\
{y_7} = 52 \\
{y_8} = 6 \times 8 + 10 \\
{y_8} = 48 + 10 \\
{y_8} = 58 \\
\]
Therefore range $ = \left\{ {16,\;22,\;28,\;34,\;40,\;46,\;52,\;58} \right\}$
Note: In solving this type of question, make sure the constant part is also written in the function, because it happens most of the times that students forget the constant part and just write the function or the equation with dependent and independent variables.
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