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1.7 - Functions. 1.7 - Functions. A function is a relation in which each element of the domain is paired with exactly one element of the range. 1.7 - Functions. A function is a relation in which each element of the domain is paired with exactly one element of the range. 1.7 - Functions.
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1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range.
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range.
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range.
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range.
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range. • There cannot be an x-value repeated!
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range. • There cannotbe an x-value repeated!
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range. • There cannotbe an x-value repeated! Ex.1 Determine if each is a function.
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range. • There cannotbe an x-value repeated! Ex.1 Determine if each is a function. • X Y -6 -4 9 -1 -6 1 1
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range. • There cannotbe an x-value repeated! Ex.1 Determine if each is a function. • X Y -6 -4 9 Y -1 -6 E 1 1 S
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range. • There cannotbe an x-value repeated! Ex.1 Determine if each is a function. • X Y b. -6 -4 9 Y -1 -6 E 1 1 S
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range. • There cannotbe an x-value repeated! Ex.1 Determine if each is a function. • X Y b. -6 -4 9 Y -1 -6 E 1 1 S
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range. • There cannotbe an x-value repeated! Ex.1 Determine if each is a function. • X Y b. -6 -4 9 Y -1 -6 E 1 1 S
1.7 - Functions • A function is a relation in which each element of the domain is paired with exactly one element of the range. • There cannotbe an x-value repeated! Ex.1 Determine if each is a function. • X Y b. -6 NOT A -4 9 Y FUNC. -1 -6 E 1 1 S
Ex. 2 If f(x) = x2 – 5, find the following: a. f(-9) f(x) = x2 – 5
Ex. 2 If f(x) = x2 – 5, find the following: a. f(-9) f(x) = x2 – 5 f(-9)
Ex. 2 If f(x) = x2 – 5, find the following: a. f(-9) f(x) = x2 – 5 f(-9)
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 – 5 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 – 5 f(6z) = 36z2 – 5 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 – 5 f(6z) = 36z2 – 5 c. f(4) + 2 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 – 5 f(6z) = 36z2 – 5 c. f(4) + 2 f(4) + 2 = Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 – 5 f(6z) = 36z2 – 5 c. f(4) + 2 f(4) + 2 = [(4)2 – 5] Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 – 5 f(6z) = 36z2 – 5 c. f(4) + 2 f(4) + 2 =[(4)2 – 5] Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 – 5 f(6z) = 36z2 – 5 c. f(4) + 2 f(4) + 2 =[(4)2 – 5] + 2 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 – 5 f(6z) = 36z2 – 5 c. f(4) + 2 f(4) + 2 = [(4)2 – 5] + 2 = [16 – 5] + 2 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 – 5 f(6z) = 36z2 – 5 c. f(4) + 2 f(4) + 2 = [(4)2 – 5] + 2 = [16 – 5] + 2 = 11 + 2 Ex. 2 If f(x) = x2 – 5, find the following:
a. f(-9) f(x) = x2 – 5 f(-9) = (-9)2 – 5 = 81 – 5 f(-9) = 76 b. f(6z) f(x) = x2 – 5 f(6z) = (6z)2 – 5 = 62·z2 – 5 f(6z) = 36z2 – 5 c. f(4) + 2 f(4) + 2 = [(4)2 – 5] + 2 = [16 – 5] + 2 = 11 + 2 f(4) + 2 = 13 Ex. 2 If f(x) = x2 – 5, find the following: