Question 461376
Using the formula {{{f(x)=x^2-2x+1}}}, find if it has a maximum or minimum and give that point. Also give x-intercepts.

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Find minimum/maximum point
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This function represents a the standard form of an equation for a parabola, {{{f(x)=ax^2+bx+c}}}. 
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There is an easy way to tell if the parabola has a maximum or a minimum when the equation is in this form. If the coefficient of the x-squared term is positive the parabola opens upward, and the the function has a minimum at the vertex. If the coefficient is negative, the parabola opens downward, and the the function has a maximum value.
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In your equation, the coefficient is 1 because you have {{{x^2}}}. (Remember, if no coefficient is shown, it is understood to be 1.) Your parabola opens upward, and the function has a minimum value.
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Parabolas have one maximum or minimum value located at the vertex. To find the vertex of your parabola, we use the formula, x = -b/2a. Looking at the equation, we see that b is -2 and a is 1, so x = -b/2a = -(-2)/(2(1)) = 1. To find the y-value of the vertex, substitute 1 for x in the equation. 
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{{{y=x^2-2x+1=(1)^2-2(1)+1)=0}}}
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Find the x-intercepts
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The x-intercepts are the points where the parabola intersects the x-axis. To find the x-intercepts we set f(x) = 0. and solve for x.
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{{{f(x)=x^2-2x+1}}}
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Set the function equal to zero.
{{{0=x^2-2x+1}}}
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Invert the order of the equation.
{{{x^2-2x+1=0}}}
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This is a quadratic equation. We have two choices for solving it: factoring, or the quadratic formula. In this case the equation looks pretty east to factor.
{{{(x-1)(x-1)=0}}}
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Notice that both factors are the same, so
{{{x-1=0}}}
{{{x=1}}}
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This tells us that when f(x) is 0, x is 1 and the x-intercept is the point (1,0).
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A quick check using substitution, gives us
{{{f(x)=x^2-2x+1}}}
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{{{0=(1)^2-2(1)+1}}}
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{{{0=1-2+1}}}
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{{{0=0}}} (true)

Take a look at the graph:
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{{{graph( 300, 300, -5, 5, -5, 5, x^2-2x+1)}}}


Hope this helps!

Ms. Figgy
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