Stitz-Zeager_College_Algebra_e-book

In a similar fashion we recall that to test an

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: h satisfy the equation y = f (x). That is, the point (x, y ) is on the graph of f if and only if y = f (x). Example 1.7.1. Graph f (x) = x2 − x − 6. Solution. To graph f , we graph the equation y = f (x). To this end, we use the techniques outlined in Section 1.3. Specifically, we check for intercepts, test for symmetry, and plot additional points as needed. To find the x-intercepts, we set y = 0. Since y = f (x), this means f (x) = 0. f (x) 0 0 x−3=0 x = = = or = x2 − x − 6 x2 − x − 6 (x − 3)(x + 2) factor x+2=0 −2, 3 So we get (−2, 0) and (3, 0) as x-intercepts. To find the y -intercept, we set x = 0. Using function notation, this is the same as finding f (0) and f (0) = 02 − 0 − 6 = −6. Thus the y -intercept is (0, −6). As far as symmetry is concerned, we can tell from the intercepts that the graph possesses none of the three symmetries discussed thus far. (You should verify this.) We can make a table analogous to the ones we made in Section 1.3, plot the points and connect the dots in a somewhat plea...
View Full Document

Ask a homework question - tutors are online