Write X^2-6x+1 in the form (x+a)^2+b?
回答 (7)
2020-09-07 2:57 pm
x² - 6x + 1
= (x² - 6x) + 1
= (x² - 2*x*3 + 3²) + 1 - 3²
= (x - 3)² - 8
= [x + (-3)]² + (-8)
= (x² - 6x) + 1
= (x² - 2*x*3 + 3²) + 1 - 3²
= (x - 3)² - 8
= [x + (-3)]² + (-8)
2020-09-09 5:48 am
x^2-6x+1
=
x^2-6x+3^2-3^2+1
=
(x-3)^2-8
=>
a=3
b=-8
=
x^2-6x+3^2-3^2+1
=
(x-3)^2-8
=>
a=3
b=-8
2020-09-07 9:56 pm
x^2 - 6x + 1
= (x - 3)^2 - 8
= (x - 3)^2 - 8
2020-09-08 4:52 am
x^2 - 6x + 1
= (x - 3)^2 - 8
= (x - 3)^2 - 8
2020-09-07 9:25 pm
Complete the square.
x^2 - 6x + 1 =
x^2 - 6x + (6/2)^2 + 1 - (6/2)^2 =
(x - 3)^2 + 1 - 9 =
(x - 3)^2 - 8
x^2 - 6x + 1 =
x^2 - 6x + (6/2)^2 + 1 - (6/2)^2 =
(x - 3)^2 + 1 - 9 =
(x - 3)^2 - 8
2020-09-07 6:26 pm
Complete the square:
x² - 6x + 1
= (x² - 6x) + 1
= (x² - 6x + 9) - 9 + 1
= (x-3)² - 8
x² - 6x + 1
= (x² - 6x) + 1
= (x² - 6x + 9) - 9 + 1
= (x-3)² - 8
2020-09-07 3:15 pm
Here's how I do it. Similar process, but I think a little more streamlined.
You are asked to put that quadratic into vertex form. I'll set the expression to a variable so we have an equation, then we need to get the right side of that equation into the form of (x² + bx). We can do that by subtracting 1 from both sides:
y = x² - 6x + 1
y - 1 = x² - 6x
Now we complete the square by adding 9 to both sides:\
y + 8 = x² - 6x + 9
The right side now factors:
y + 8 = (x - 3)²
Now solve for y again:
y = (x - 3)² - 8
You are asked to put that quadratic into vertex form. I'll set the expression to a variable so we have an equation, then we need to get the right side of that equation into the form of (x² + bx). We can do that by subtracting 1 from both sides:
y = x² - 6x + 1
y - 1 = x² - 6x
Now we complete the square by adding 9 to both sides:\
y + 8 = x² - 6x + 9
The right side now factors:
y + 8 = (x - 3)²
Now solve for y again:
y = (x - 3)² - 8
收錄日期: 2021-05-01 22:45:10
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