What is y=3(x-1)^2+5 in standard form?
回答 (4)
2009-06-17 3:45 pm
✔ 最佳答案
standard form is y = ax^2 + bx + cso expanding, y = 3(x^2 - 2x + 1) + 5
= 3x^2 - 6x + 3 + 5
= 3x^2 - 6x + 8
so y = 3x^2 - 6x + 8
2009-06-17 10:46 pm
"Confused" is correct. Follow his/her steps.
2009-06-17 10:45 pm
3(x-1)²+5
3(x²-2x+1)+5
3x²-6x+3+5
3x²-6x+8
3(x²-2x+1)+5
3x²-6x+3+5
3x²-6x+8
2009-06-17 10:55 pm
the structure of standard form is y=ax^2+bx+c
step 1: expand (x-1)^2
y=3(x-1)^2+5
y=3(x^2-2x+1)+5
step 2: distribute the constant "3" to (x^2-2x+1)
y=3x^2-6x+3+5
step 3: combining like terms
y=3x^2-6x+8
step 1: expand (x-1)^2
y=3(x-1)^2+5
y=3(x^2-2x+1)+5
step 2: distribute the constant "3" to (x^2-2x+1)
y=3x^2-6x+3+5
step 3: combining like terms
y=3x^2-6x+8
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