You can put this solution on YOUR website! We use Newton's method to approximate zeros
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x(n+1) = x(n) - (f(x(n)) / f'(x(n)))
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1) y = x^2-7x+7
Consider the graph of this equation
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The zeros are close to 1 and 6, I will show you how to find the zero close to 1. All others follow the same procedure.
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Calculate f'
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f' = 2x-7
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Let x(0) = 1, then
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x(1) = x(0) - (f(x(0)) / f'(x(0))
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x(1) = 1 - (1 / -5) = 1.2
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x(2) = 1.2 - (0.04 / -4.6) approx 1.2
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Since x(2) = x(1), there is no point in continuing
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f(1.2) = 0.04 approximately 0.0, therefore
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x = 1.2 is a good approximation for a zero
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