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Solution 2.3:6a

From Förberedande kurs i matematik 1

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Current revision (11:29, 29 September 2008) (edit) (undo)
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Using the squaring rule, we recognize the polynomial as the expansion of
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Using the rule <math>(a+b)^2=a^2+2ab+b^2</math>, we recognize the polynomial as the expansion of <math>(x-1)^{2}\,</math>,
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<math>\left( x-1 \right)^{2}</math>,
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{{Displayed math||<math>x^{2}-2x+1 = (x-1)^{2}\,\textrm{.}</math>}}
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<math>x^{2}-2x+1=\left( x-1 \right)^{2}</math>
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This quadratic expression has its smallest value, zero, when <math>x-1=0</math>, i.e.
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<math>x=1</math>. All non-zero values of <math>x-1</math> give a positive value for
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<math>(x-1)^{2}</math>.
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This quadratic expression has its smallest value, zero, when
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Note: If we draw the curve <math>y=(x-1)^{2}</math>, we see that it has a minimum value of zero at <math>x=1\,</math>.
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<math>x-\text{1}=0</math>, i.e.
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<math>x=\text{1}</math>. All non-zero values of
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<math>x-\text{1}</math>
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give a positive value for
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<math>\left( x-1 \right)^{2}</math>.
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NOTE: If we draw the curve
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<math>y=\left( x-1 \right)^{2}</math>, we see that it has a minimum value of zero at
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<math>x=\text{1}</math>.
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[[Image:2_3_6_a.gif|center]]
[[Image:2_3_6_a.gif|center]]

Current revision

Using the rule (a+b)2=a2+2ab+b2, we recognize the polynomial as the expansion of (x−1)2,

x2−2x+1=(x−1)2.

This quadratic expression has its smallest value, zero, when x−1=0, i.e. x=1. All non-zero values of x−1 give a positive value for (x−1)2.


Note: If we draw the curve y=(x−1)2, we see that it has a minimum value of zero at x=1.


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