Math Problem Statement

For the following exercises, evaluate the function at the indicated values: f(−3); f(2); f(−a); −f(a); f(a+h)

.

f(x)=−2x2+3x

f(x)=2|3x−1|

Solution

Let's evaluate the functions step by step at the indicated values:


Exercise 5: f(x)=2x2+3xf(x) = -2x^2 + 3x

(a) f(3)f(-3):
Substitute x=3x = -3 into f(x)f(x): f(3)=2(3)2+3(3)f(-3) = -2(-3)^2 + 3(-3) f(3)=2(9)9=189=27f(-3) = -2(9) - 9 = -18 - 9 = -27


(b) f(2)f(2):
Substitute x=2x = 2 into f(x)f(x): f(2)=2(2)2+3(2)f(2) = -2(2)^2 + 3(2) f(2)=2(4)+6=8+6=2f(2) = -2(4) + 6 = -8 + 6 = -2


(c) f(a)f(-a):
Substitute x=ax = -a into f(x)f(x): f(a)=2(a)2+3(a)f(-a) = -2(-a)^2 + 3(-a) f(a)=2(a2)3a=2a23af(-a) = -2(a^2) - 3a = -2a^2 - 3a


(d) f(a)-f(a):
First find f(a)f(a) by substituting x=ax = a into f(x)f(x): f(a)=2a2+3af(a) = -2a^2 + 3a Now multiply by 1-1: f(a)=(2a2+3a)=2a23a-f(a) = -(-2a^2 + 3a) = 2a^2 - 3a


(e) f(a+h)f(a+h):
Substitute x=a+hx = a+h into f(x)f(x): f(a+h)=2(a+h)2+3(a+h)f(a+h) = -2(a+h)^2 + 3(a+h) First expand (a+h)2(a+h)^2: (a+h)2=a2+2ah+h2(a+h)^2 = a^2 + 2ah + h^2 Now substitute: f(a+h)=2(a2+2ah+h2)+3(a+h)f(a+h) = -2(a^2 + 2ah + h^2) + 3(a+h) Distribute: f(a+h)=2a24ah2h2+3a+3hf(a+h) = -2a^2 - 4ah - 2h^2 + 3a + 3h Thus: f(a+h)=2a24ah2h2+3a+3hf(a+h) = -2a^2 - 4ah - 2h^2 + 3a + 3h


Exercise 6: f(x)=23x1f(x) = 2|3x - 1|

(a) f(3)f(-3):
Substitute x=3x = -3 into f(x)f(x): f(3)=23(3)1f(-3) = 2|3(-3) - 1| f(3)=291=210=2(10)=20f(-3) = 2| -9 - 1| = 2| -10 | = 2(10) = 20


(b) f(2)f(2):
Substitute x=2x = 2 into f(x)f(x): f(2)=23(2)1f(2) = 2|3(2) - 1| f(2)=261=25=2(5)=10f(2) = 2|6 - 1| = 2|5| = 2(5) = 10


(c) f(a)f(-a):
Substitute x=ax = -a into f(x)f(x): f(a)=23(a)1f(-a) = 2|3(-a) - 1| f(a)=23a1f(-a) = 2| -3a - 1 |


(d) f(a)-f(a):
First find f(a)f(a) by substituting x=ax = a into f(x)f(x): f(a)=23a1f(a) = 2|3a - 1| Now multiply by 1-1: f(a)=23a1-f(a) = -2|3a - 1|


(e) f(a+h)f(a+h):
Substitute x=a+hx = a+h into f(x)f(x): f(a+h)=23(a+h)1f(a+h) = 2|3(a+h) - 1| Simplify inside the absolute value: f(a+h)=23a+3h1f(a+h) = 2|3a + 3h - 1|


Final Answers Summary:

Exercise 5: f(x)=2x2+3xf(x) = -2x^2 + 3x

  1. f(3)=27f(-3) = -27
  2. f(2)=2f(2) = -2
  3. f(a)=2a23af(-a) = -2a^2 - 3a
  4. f(a)=2a23a-f(a) = 2a^2 - 3a
  5. f(a+h)=2a24ah2h2+3a+3hf(a+h) = -2a^2 - 4ah - 2h^2 + 3a + 3h

Exercise 6: f(x)=23x1f(x) = 2|3x - 1|

  1. f(3)=20f(-3) = 20
  2. f(2)=10f(2) = 10
  3. f(a)=23a1f(-a) = 2| -3a - 1 |
  4. f(a)=23a1-f(a) = -2|3a - 1|
  5. f(a+h)=23a+3h1f(a+h) = 2|3a + 3h - 1|

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Related Questions:

  1. How do you simplify expressions involving absolute values like 3x1|3x - 1|?
  2. What are the properties of quadratic functions like 2x2+3x-2x^2 + 3x?
  3. How do you expand (a+h)2(a+h)^2 step by step?
  4. How does multiplying by 1-1 affect a function like f(a)-f(a)?
  5. How do you evaluate composite functions, such as f(g(x))f(g(x))?

Tip: Always distribute carefully when expanding squared binomials and keep track of signs!

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Absolute Value Functions
Function Evaluation
Binomial Expansion

Formulas

f(x) = −2x^2 + 3x
f(x) = 2|3x−1|
Binomial expansion: (a + b)^2 = a^2 + 2ab + b^2

Theorems

Properties of Absolute Value
Distributive Property
Exponentiation Rules

Suitable Grade Level

Grades 8-10