What is the axis y=+1/7 x^2 ?

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axis y=+1/7 x^2

axis

y = + \frac{1}{7} x^{2}

x = 0

Solution steps

y = \frac{1}{7} x^{2}

Rewrite

y = \frac{1}{7} x^{2}

in the standard form:

4 \cdot \frac{7}{4} (y - 0) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{7}{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 = - 8 x^{2} + 4

x = + 2 (y + 1)^{2} - 5

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

y = 3.1 + 6.6 x - 2.3 x^{2}

y = 2 x^{2} + 4 x - 3