Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Problem Statement:**

Put the equation \( y = -3(x + 2)^2 + 5 \) into general form \( y = ax^2 + bx + c \).

**Solution:**

To convert the given equation into general form, we need to expand it:

1. Start with the equation:
   \[
   y = -3(x + 2)^2 + 5
   \]

2. Expand \( (x + 2)^2 \):
   \[
   (x + 2)^2 = x^2 + 4x + 4
   \]

3. Substitute back into the equation:
   \[
   y = -3(x^2 + 4x + 4) + 5
   \]

4. Distribute the \(-3\):
   \[
   y = -3x^2 - 12x - 12 + 5
   \]

5. Combine like terms:
   \[
   y = -3x^2 - 12x - 7
   \]

The equation in general form is \( y = -3x^2 - 12x - 7 \).
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Transcribed Image Text:**Problem Statement:** Put the equation \( y = -3(x + 2)^2 + 5 \) into general form \( y = ax^2 + bx + c \). **Solution:** To convert the given equation into general form, we need to expand it: 1. Start with the equation: \[ y = -3(x + 2)^2 + 5 \] 2. Expand \( (x + 2)^2 \): \[ (x + 2)^2 = x^2 + 4x + 4 \] 3. Substitute back into the equation: \[ y = -3(x^2 + 4x + 4) + 5 \] 4. Distribute the \(-3\): \[ y = -3x^2 - 12x - 12 + 5 \] 5. Combine like terms: \[ y = -3x^2 - 12x - 7 \] The equation in general form is \( y = -3x^2 - 12x - 7 \).
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