What is the axis y= 1/2 (x^2+1)-3/2 (x^2-1) ?

The axis y= 1/2 (x^2+1)-3/2 (x^2-1) is x=0

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axis y= 1/2 (x^2+1)-3/2 (x^2-1)

axis

y = \frac{1}{2} (x^{2} + 1) - \frac{3}{2} (x^{2} - 1)

x = 0

Solution steps

y = \frac{1}{2} (x^{2} + 1) - \frac{3}{2} (x^{2} - 1)

Rewrite

y = \frac{1}{2} (x^{2} + 1) - \frac{3}{2} (x^{2} - 1)

in the standard form

4 (- \frac{1}{4}) (y - 2) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 2), p = - \frac{1}{4}

Parabola is of the form

4 p (y - k) = (x - h)^{2}

and is symmetric around the

y

-axis

Axis of symmetry is a line parallel to the

y

-axis which intersects the vertex:

x = 0

y = - 2 x^{2} + 5 x - 6

y^{2} - 3 y - x + 3 = 0

x^{2} - x + 5

y = 2 (x - 9)^{2}

y = s^{2} + s - 2

What is the axis y= 1/2 (x^2+1)-3/2 (x^2-1) ?
The axis y= 1/2 (x^2+1)-3/2 (x^2-1) is x=0