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find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=-2x^2-8x+3. explain the steps please. :)
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Answered by
MathMate
The axis of symmetry of a parabola of the form y=ax²+bx+c is always vertical.
If a > 0, the parabola is concave upwards, and if a<0, the parabola is concave downwards.
The vertex of a parabola can be obtained by completing the squares, namely
y=a((x+(b/2a)²+c/a-(b/2a)²)
or at
x= -b/2a, at which point
y=c-b²/4a
Thus the vertex of the parabola
y=ax²+bx+c is at
(-b/2a, c-b²/4a)
The vertical axis of symmetry passes through the vertex, thus has the equation:
x=-b/2a
Apply the above the equation to the given problem and post the answer for verification if required.
If a > 0, the parabola is concave upwards, and if a<0, the parabola is concave downwards.
The vertex of a parabola can be obtained by completing the squares, namely
y=a((x+(b/2a)²+c/a-(b/2a)²)
or at
x= -b/2a, at which point
y=c-b²/4a
Thus the vertex of the parabola
y=ax²+bx+c is at
(-b/2a, c-b²/4a)
The vertical axis of symmetry passes through the vertex, thus has the equation:
x=-b/2a
Apply the above the equation to the given problem and post the answer for verification if required.
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