Write a function that describes a relationship between two quantities - HiSET

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is a linear function defined on the set of real numbers. Four values of are given in the following table:

$\begin{matrix} $\underline{x}$& $\underline{f(x)}$ \ 13 & 4 \ 10 & 6 \ 7 & 8 \ 4 & 10 \end{matrix}$

Give the definition for .

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By the two-point formula, the equation of a line through points is

,

where $m = $$\frac{x_{2}$$$-x_{1}$$}{y_{2}$$-y_{1}$}$ , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting , then

$m = $$\frac{x_{2}$$$-x_{1}$$}{y_{2}$$-y_{1}$}= $\frac{7-10}{8-6}$ = -$\frac{3}{2}$$

The first formula becomes

$y-10 = -$\frac{3}{2}$(x-6)$

Since , solve for . First, distribute $-$\frac{3}{2}$$ throughout the difference at right:

$y-10 = -$\frac{3}{2}$ x- \left (-$\frac{3}{2}$ \right )6$

$y-10 = -$\frac{3}{2}$ x- \left (-9 \right )$

$y-10 = -$\frac{3}{2}$ x+9$

Isolate by adding 10 to both sides:

$y-10+ 10 = -$\frac{3}{2}$ x+9+ 10$

$y = -$\frac{3}{2}$ x+19$

Replacing, the definition of is

$f(x) = -$\frac{3}{2}$ x+19$ .

is a linear function defined on the set of real numbers. Four values of are given in the following table:

$\begin{matrix} $\underline{x}$& $\underline{f(x)}$ \ 13 & 4 \ 10 & 6 \ 7 & 8 \ 4 & 10 \end{matrix}$

Give the definition for .

Tap to see back →

By the two-point formula, the equation of a line through points is

,

where $m = $$\frac{x_{2}$$$-x_{1}$$}{y_{2}$$-y_{1}$}$ , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting , then

$m = $$\frac{x_{2}$$$-x_{1}$$}{y_{2}$$-y_{1}$}= $\frac{7-10}{8-6}$ = -$\frac{3}{2}$$

The first formula becomes

$y-10 = -$\frac{3}{2}$(x-6)$

Since , solve for . First, distribute $-$\frac{3}{2}$$ throughout the difference at right:

$y-10 = -$\frac{3}{2}$ x- \left (-$\frac{3}{2}$ \right )6$

$y-10 = -$\frac{3}{2}$ x- \left (-9 \right )$

$y-10 = -$\frac{3}{2}$ x+9$

Isolate by adding 10 to both sides:

$y-10+ 10 = -$\frac{3}{2}$ x+9+ 10$

$y = -$\frac{3}{2}$ x+19$

Replacing, the definition of is

$f(x) = -$\frac{3}{2}$ x+19$ .

is a linear function defined on the set of real numbers. Four values of are given in the following table:

$\begin{matrix} $\underline{x}$& $\underline{f(x)}$ \ 13 & 4 \ 10 & 6 \ 7 & 8 \ 4 & 10 \end{matrix}$

Give the definition for .

Tap to see back →

By the two-point formula, the equation of a line through points is

,

where $m = $$\frac{x_{2}$$$-x_{1}$$}{y_{2}$$-y_{1}$}$ , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting , then

$m = $$\frac{x_{2}$$$-x_{1}$$}{y_{2}$$-y_{1}$}= $\frac{7-10}{8-6}$ = -$\frac{3}{2}$$

The first formula becomes

$y-10 = -$\frac{3}{2}$(x-6)$

Since , solve for . First, distribute $-$\frac{3}{2}$$ throughout the difference at right:

$y-10 = -$\frac{3}{2}$ x- \left (-$\frac{3}{2}$ \right )6$

$y-10 = -$\frac{3}{2}$ x- \left (-9 \right )$

$y-10 = -$\frac{3}{2}$ x+9$

Isolate by adding 10 to both sides:

$y-10+ 10 = -$\frac{3}{2}$ x+9+ 10$

$y = -$\frac{3}{2}$ x+19$

Replacing, the definition of is

$f(x) = -$\frac{3}{2}$ x+19$ .

is a linear function defined on the set of real numbers. Four values of are given in the following table:

$\begin{matrix} $\underline{x}$& $\underline{f(x)}$ \ 13 & 4 \ 10 & 6 \ 7 & 8 \ 4 & 10 \end{matrix}$

Give the definition for .

Tap to see back →

By the two-point formula, the equation of a line through points is

,

where $m = $$\frac{x_{2}$$$-x_{1}$$}{y_{2}$$-y_{1}$}$ , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting , then

$m = $$\frac{x_{2}$$$-x_{1}$$}{y_{2}$$-y_{1}$}= $\frac{7-10}{8-6}$ = -$\frac{3}{2}$$

The first formula becomes

$y-10 = -$\frac{3}{2}$(x-6)$

Since , solve for . First, distribute $-$\frac{3}{2}$$ throughout the difference at right:

$y-10 = -$\frac{3}{2}$ x- \left (-$\frac{3}{2}$ \right )6$

$y-10 = -$\frac{3}{2}$ x- \left (-9 \right )$

$y-10 = -$\frac{3}{2}$ x+9$

Isolate by adding 10 to both sides:

$y-10+ 10 = -$\frac{3}{2}$ x+9+ 10$

$y = -$\frac{3}{2}$ x+19$

Replacing, the definition of is

$f(x) = -$\frac{3}{2}$ x+19$ .