Exercise 65 Page 320

Equations in point-slope form follow a specific format.

y - y_{1} = m (x - x_{1})

In this form,

m

is the slope of the line and

(x_{1}, y_{1})

is a point on the line. We are looking for a function with values

f (- 2) = 0

and

f (6) = - 4 .

This means that the linear function passes through two known points.

(- 2, 0) and (6, - 4)

To determine the slope of the line, we use the Slope Formula.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Substitute $(- 2, 0)$ & $(6, - 4)$

m = \frac{- 4 - 0}{6 - ( - 2 )}

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Simplify right-hand side

$a - (- b) = a + b$

m = \frac{- 4 - 0}{6 + 2}

Add and subtract terms

m = \frac{- 4}{8}

$\frac{a}{b} = \frac{a / 4}{b / 4}$

m = \frac{- 1}{2}

Put minus sign in front of fraction

m = - \frac{1}{2}

Now that we know the slope of the line is

- \frac{1}{2},

we can write the equation of the line in point-slope form. We can use either of the given points as

(x_{1}, y_{1})

in our equation. Let's use

(- 2, 0) .

y - y_{1} = m (x - x_{1})

Substitute values

y - 0 = - \frac{1}{2} (x - (- 2))

$a - (- b) = a + b$

y - 0 = - \frac{1}{2} (x + 2)

Distribute $- \frac{1}{2}$

y - 0 = - \frac{1}{2} x - 1

Identity Property of Addition

y = - \frac{1}{2} x - 1

We found a line

y = - \frac{1}{2} x - 1,

so the function we are asked for is

f (x) = - \frac{1}{2} x - 1 .