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Put the equation y = y = - 3(x + 2)² + 5 into general form y = ax² + bx + c.

Put the equation y = y = - 3(x + 2)² + 5 into general form y = ax² + bx + c.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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 \( y = -3x^2 - 12x - 7 \).
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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