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Let ๐‘“โก(๐‘ฅ)= -3โข๐‘ฅ2-2. Find and simplify the following:(a) ๐‘“โข(๐‘ฅ-5)=  (b) ๐‘“โก(๐‘ฅ-5)-๐‘“โก(-5)=

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Solution

1. Break Down the Problem

We have the function f(x)=โˆ’3x2โˆ’2 f(x) = -3x^2 - 2 . We need to solve two parts:

  1. Find f(xโˆ’5) f(x - 5) .
  2. Calculate f(xโˆ’5)โˆ’f(โˆ’5) f(x - 5) - f(-5) .

2. Relevant Concepts

The function f(x)=โˆ’3x2โˆ’2 f(x) = -3x^2 - 2 is a quadratic function. We will substitute xโˆ’5 x - 5 and โˆ’5-5 into the function to get the desired values.

3. Analysis and Detail

Part (a) f(xโˆ’5) f(x - 5)

To find f(xโˆ’5) f(x - 5) : f(xโˆ’5)=โˆ’3(xโˆ’5)2โˆ’2 f(x-5) = -3(x-5)^2 - 2 Calculating (xโˆ’5)2 (x-5)^2 : (xโˆ’5)2=x2โˆ’10x+25 (x-5)^2 = x^2 - 10x + 25 Then substituting back: f(xโˆ’5)=โˆ’3(x2โˆ’10x+25)โˆ’2 f(x-5) = -3(x^2 - 10x + 25) - 2 Distributing โˆ’3-3: f(xโˆ’5)=โˆ’3x2+30xโˆ’75โˆ’2 f(x-5) = -3x^2 + 30x - 75 - 2 Combining like terms: f(xโˆ’5)=โˆ’3x2+30xโˆ’77 f(x-5) = -3x^2 + 30x - 77

Part (b) f(xโˆ’5)โˆ’f(โˆ’5) f(x - 5) - f(-5)

First, we need to compute f(โˆ’5) f(-5) : f(โˆ’5)=โˆ’3(โˆ’5)2โˆ’2=โˆ’3(25)โˆ’2=โˆ’75โˆ’2=โˆ’77 f(-5) = -3(-5)^2 - 2 = -3(25) - 2 = -75 - 2 = -77 Now, substituting both results: f(xโˆ’5)โˆ’f(โˆ’5)=(โˆ’3x2+30xโˆ’77)โˆ’(โˆ’77) f(x - 5) - f(-5) = (-3x^2 + 30x - 77) - (-77) This simplifies to: f(xโˆ’5)โˆ’f(โˆ’5)=โˆ’3x2+30xโˆ’77+77 f(x - 5) - f(-5) = -3x^2 + 30x - 77 + 77 Thus: f(xโˆ’5)โˆ’f(โˆ’5)=โˆ’3x2+30x f(x - 5) - f(-5) = -3x^2 + 30x

4. Verify and Summarize

We derived f(xโˆ’5)=โˆ’3x2+30xโˆ’77 f(x - 5) = -3x^2 + 30x - 77 and f(โˆ’5)=โˆ’77 f(-5) = -77 . Finally, we computed f(xโˆ’5)โˆ’f(โˆ’5)=โˆ’3x2+30x f(x - 5) - f(-5) = -3x^2 + 30x successfully.

Final Answer

(a) f(xโˆ’5)=โˆ’3x2+30xโˆ’77 f(x - 5) = -3x^2 + 30x - 77
(b) f(xโˆ’5)โˆ’f(โˆ’5)=โˆ’3x2+30x f(x - 5) - f(-5) = -3x^2 + 30x

This problem has been solved

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