Question 110039
In a quadratic equation of the standard form:
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{{{ax^2 + bx + c = 0}}}
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the axis of symmetry will be given by the equation:
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{{{x = -b/(2a)}}}
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Note that this equation for the axis of symmetry is the first term of the solution for x that is
provided by the quadratic formula:
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{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
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In this problem you are given the quadratic equation:
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{{{y=x^2-10x+7}}}
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You can convert this to the standard quadratic form by setting y equal to zero to get:
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{{{0=x^2-10x+7}}}
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and then transposing the equation (reversing sides) to change it to:
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{{{x^2-10x+7 = 0}}}
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By comparing this equation, term by term, with the standard form of {{{ax^2 + bx + c = 0}}}
you can see that "a" [the multiplier of the {{{x^2}}}] must be 1, "b" [the multiplier
of the {{{x}}} ] must be -10, and "c" [the constant] must be +7.
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Now that the values of a, b, and c have been found by comparing the standard form of
the quadratic equation to the form given in the problem, the line of symmetry can be
found by substituting these values into the equation:
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{{{x = -b/(2*a)}}}
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"a" was found to be 1 and "b" was found to be -10. Substituting these values results
in the line of symmetry equation becoming:
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{{{x = -(-10)/(2*1)}}}
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and this simplifies to:
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{{{x = 10/2 = 5}}}
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the equation for the line of symmetry is, therefore, x = +5 and the graph of this line is 
a vertical line through the point +5 on the x-axis because the equation tells you that
no matter what the value of y, x is always +5.  So all the following points are on the line:
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(5, 10) (5, 5) (5, 0) (5, -5) (5, -10) and so on
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If you plot all of these points on a graph you will see that they lie on the vertical
line through +5 on the x-axis.
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I hope this helps you to see how to work this problem.
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