What is the axis y=5x^2-7 ?

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axis y=5x^2-7

axis

y = 5 x^{2} - 7

x = 0

Solution steps

y = 5 x^{2} - 7

Rewrite

y = 5 x^{2} - 7

in the standard form:

4 \cdot \frac{1}{20} (y - (- 7)) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, - 7), p = \frac{1}{20}

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 x + 5

x = (y + 2)^{2}

y = + x^{2} + 14 x + 45

y = 1.4 x^{2} + 2 x + 1

y = 2 x^{2} - 12 x - 5