What is the axis y=4x^2+6 ?

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axis y=4x^2+6

axis

y = 4 x^{2} + 6

x = 0

Solution steps

y = 4 x^{2} + 6

Rewrite

y = 4 x^{2} + 6

in the standard form:

4 \cdot \frac{1}{16} (y - 6) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 6), p = \frac{1}{16}

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 = \frac{x ^{2} - 2 x - 19}{5}

y = x^{2} + 8 x - 1

x^{2} + y = 4

y = - x (x + 6)

- x^{2} - 10 x - 25