SOLUTION: Name the vertex and axis of symmetry for each quadratic function. Tell wherther the parabola opens up, down, left, or right. x=2y^2+4y+5

Question 194519: Name the vertex and axis of symmetry for each quadratic function. Tell wherther the parabola opens up, down, left, or right.
x=2y^2+4y+5
Found 2 solutions by RAY100, stanbon:
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
x = 2y^2 +4y +5
,
Leading coefficient is positive therefore it opens to right
,
y vertex is at -b/2a,,, -(4)/2(2) = (-1)
x vertex is found by substituting, x= 2(-1)^2 +4(-1) +5 = +3
Vertex is (3,-1)
axis of symmetry is y= -1
,
checking by plotting a few points,,,(3,-1),,,(5,0),,,,(5,-2),,,,ok
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Name the vertex and axis of symmetry for each quadratic function. Tell wherther the parabola opens up, down, left, or right.
x=2y^2+4y+5
--------------
Since the y^2 term is positive the parabola opens to the right.
-------------------
Complete the square:
2(y^2 + 2y + 1) = x - 5 + 2
2(y+1)^2 = x-3
Vertex: (3,-1)
Axis: y = -1
=====================
Cheers,
Stan H.
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