Exercise 25 Page 326

To determine if the graph of the given equation is a horizontal or vertical line, let's first consider how the given equation would be written in standard form, Ax+ By= C. y=7/4 ⇔ 0x+ 1y= 7/4Normally, when graphing a line using the standard form, we would substitute 0 for x and solve for y. Then, we would substitute 0 for y and solve for x. Let's try this method using our standard form equation. We will start with the y-intercept.

0x+1y=7/4

Substitute

x= 0

0( 0)+1y? =7/4

ZeroPropMult

Zero Property of Multiplication

0+1y? =7/4

IdPropAdd

Identity Property of Addition

1y? =7/4

IdPropMult

Identity Property of Multiplication

y=7/4

Everything is going smoothly so far. The y-intercept lies on the point (0, 74). Now let's try the x-intercept.

0x+1y=7/4

Substitute

y= 0

0x+1( 0)? =7/4

ZeroPropMult

Zero Property of Multiplication

0+0? =7/4

AddTerms

Add terms

0≠7/4

When we try to find the x-intercept of the given equation, we create a contradiction. This means that the line does not cross the x-axis, so it must be horizontal. Note that any time we have an equation in the form y=b, where b is a real number, the graph of y=b will be a horizontal line.