Question

A law of physics. Refer to exercise $4.26$. The linear equation is $y=64-32x$.

a. Find the $y$-intercept and slope of the specified linear equation.

b. Explain what the $y$-intercept and slope represent in terms of the graph of the equation.

c. Explain what the $y$-intercept and slope represent in terms relating to the application.

Short answer

(a) The $y$-intercept is, $64$and the slope is, $-32$

(b) The intersection with the vertical axis, steepens the graph.

(c) The $y$-intercept is the speed after $0s$is $64\raisebox{1ex}{$feet$}\!\left/ \!\raisebox{-1ex}{$second$}\right.$and slope states that speed decreases by $32\raisebox{1ex}{$feet$}\!\left/ \!\raisebox{-1ex}{$second$}\right.$

Step-by-step solution

Part (a) Step 1: Given information

We are given that,

Linear equation is,$y=64-32x$

Part (a) Step 2: Simplify 

From the given equation we interpret that,

The $y$-intercept is constant i.e. $64$

And the slope is the coefficient of $x$i.e. $-32$

Part (b) Step 1: Given information

We need to find the $y$-intercept and slope in terms of the graph.

Part (b) Step 2: Explanation

The $y$-intercept is an intersection of the graph of the given linear equation and vertical axis $y$, hence its value is $64$.

The slope represents the steepness of the graph, hence in this $y$decreases by $32$for every unit increase in $x$.

Part (c) Step 1: Given information 

We need to find the applications of y-intercept and slope.

Part (c) Step 2: Explanation

The $y$-intercept speeds up after $0s$and hence it is $64\raisebox{1ex}{$feet$}\!\left/ \!\raisebox{-1ex}{$second$}\right.$.

The slope states that the speed drops to $32\raisebox{1ex}{$feet$}\!\left/ \!\raisebox{-1ex}{$second$}\right.$for every second.