Question 819407
Original equation:
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x = 4y^2 + 8y
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the above quadratic equation represents a parabola that opens to the right, but is not a function in the strict sense because for each x there are two values of y 
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however, the original equation is similar to the function: f(x) = 4x^2 + 8x
and so the original equation can be analyzed by analyzing f(x)
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the above quadratic function is in standard form, with a=4, b=8, and c=0
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to solve the quadratic function, by using the quadratic formula, plug this:
4 8 0
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the two real roots (i.e. the two x-intercepts), of the quadratic are:
x = 0
x = -2
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the vertex is a minimum at: ( -1, -4 )
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the axis of symmetry is: x = -1
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domain is: all real x
range is: y > -4
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Now back to the original equation:
flip x and y
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the two real roots (i.e. the two y-intercepts), of the quadratic are:
y = 0
y = -2
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the vertex is at: ( -4, -1 )
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the axis of symmetry is: y = -1
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domain is: all real y
range is: x > -4
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