Line through a!ß #b and a"ß !b: y œ #x # Line through a"ß "b and a$ß "b: y œ "ģ0. (b) Line through a#ß "b and a!ß !b: y œ "# x (b) Line through a"ß !b and a!ß $b: m œ Line through a!ß $b and a#ß "b: m œ f(x) œ œ Section 1.3 Functions and Their Graphs 23.Ģ7. Neither graph passes the vertical line test (a) Neither graph passes the vertical line test (a)Ģ2. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ ˆx, Èx‰ œ ˆ m" ,Ģ1. Let D œ diagonal of a face of the cube and j œ the length of an edge.
base œ x (height)# ˆ #x ‰ œ x# Ê height œ (b) No division by ! undefined (d) Ð!ß "Óġ0. (b) Not the graph of a function of x since it fails the vertical line test. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. domain œ ('ß 2) from Exercise 5 smallest value is g(0) œ "# and as 0 z increases to 2, g(z) gets larger and larger (also true as z 0 decreases to ') Ê range œ "# ß _‰. Largest value is g(0) œ È4 œ 2 and smallest value is g(') œ g(2) œ È0 œ 0 Ê range œ. If t œ 0, then y œ 1 and as t increases, y becomes a smallerĪnd smaller positive real number Ê range œ (0ß 1]. Section 1.3 Functions and Their Graphs 4. domain œ (!ß _) y in range Ê y œĪnd y ! Ê y can be any positive real numberĬopyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley ,b Thus, È(?x)# (?y)# œ É aAx A+y œ +ġ.3 FUNCTIONS AND THEIR GRAPHS 1. Solving these equations simultaneously we find the point of intersection Q(xß y) with x œ Thus, L: Ax By œ C and M: +x Ay œ Bx! Ay!. If both A and B are Á 0 then L has slope AB so M has slope AB. Similarly, if A œ 0 and B Á 0, the distance is ¸ CB y! ¸. (d) If B œ 0 and A Á 0, then the distance from P to L is ¸ AC x! ¸ as in (c). Thus, the distance from P to L is È(a 1)# 0# œ ka 1k. The intersection point of L and M is Q("ß b). (c) M is a horizontal line with equation y œ b. Substitution 12 ‰ ˆ 12 84 ‰ into either equation gives x œ 43 ˆ 84 25 4 œ 25 so that Q 25 ß 25 is the point of intersection. Adding these we get 25 12 y œ 7 so y œ 25. We can rewrite the equations ofĨ4 the lines as L: x y œ 3 and M: + 43 y œ 4. The distance from P to L œ the distance from P to Q œ Ɉ #3 ‰# ˆ 3# ‰# œ É 18 4 œ (b) L has slope 43 so M has slopeĪnd M has the equation 4y 3x œ 12. At the intersection point, Q, we have equal y-values, y œ x 2 œ x 3. (a) L has slope 1 so M is the line through P(2ß 1) with slope ! or the line y œ x 3. Since the two triangles shown in the figure are congruent, the value a must lie midway between x" and x#, so a œ x #x. The line through (1ß 1) and ("ß #) is vertical with equation x œ 1. Substitution into either equation gives x œ 1 Ê (1ß 1) is the intersection point. Subtracting these equations we find 7y œ 7 or y œ 1.
At the point of intersection, 2x 4y œ 6 and 2x 3y œ !.
The lines are perpendicular when 2k ($) œ ! or k œ ( and parallel when 2k œ $ or k œ "#. 2x ky œ 3 has slope 2k and 4x y œ 1 has slope $.