What is the axis y=7x^2-8 ?

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axis y=7x^2-8

axis

y = 7 x^{2} - 8

x = 0

Solution steps

y = 7 x^{2} - 8

Rewrite

y = 7 x^{2} - 8

in the standard form:

4 \cdot \frac{1}{28} (y - (- 8)) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, - 8), p = \frac{1}{28}

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

y = 3 x^{2} + 5 x

y = 3 - 5 (x + 1)^{2}

2 y^{2} = - 3 x

y = 3 x^{2} + 6 x - 1