Quadratic equations are not nol in the form f(x) = ax²+bx+c to find (h, k). Begin by completing the square on a(x-h)² + k. We can complete the square on a the terms containing x² and x. EXAMPLE 6-3 Find the vertex. · y = x² - 6x-2 y = x² - 6x-2 y = |x² - 6x + (99] 2 - 2 + ? We need to balance putting +()² = 9 inside the parentheses by adding -9 outside the parentheses. y = (x² - 6x + 9) - 2-9 y = (x - 3)² - 11 The vertex is (3,-11).

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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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Can you help me by breaking down the algebra involved in getting 6/2^2 in the parentheses and -2 outside of it? I don't know what rule I'm missing to not know how this works and how to apply it to similar problems.

in the
Quadratic equations are not
form f(x) = ax²+bx+c to find (h, k). Begin by completing the square on
a(x-h)² + k. We can complete the square on a
the terms containing x² and x.
EXAMPLE 6-3
Find the vertex.
• y = x² - 6x - 2
y = x² - 6x - 2
y =
x² - 6x +
6
01-
2
- 2 + ?
We need to balance putting +()² = 9 inside the parentheses by adding
-9 outside the parentheses.
y = (x² - 6x + 9) - 2-9
y = (x - 3)² - 11
The vertex is (3, -11).
Transcribed Image Text:in the Quadratic equations are not form f(x) = ax²+bx+c to find (h, k). Begin by completing the square on a(x-h)² + k. We can complete the square on a the terms containing x² and x. EXAMPLE 6-3 Find the vertex. • y = x² - 6x - 2 y = x² - 6x - 2 y = x² - 6x + 6 01- 2 - 2 + ? We need to balance putting +()² = 9 inside the parentheses by adding -9 outside the parentheses. y = (x² - 6x + 9) - 2-9 y = (x - 3)² - 11 The vertex is (3, -11).
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