What is the axis y=5x^2+4 ?

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axis y=5x^2+4

axis

y = 5 x^{2} + 4

x = 0

Solution steps

y = 5 x^{2} + 4

Rewrite

y = 5 x^{2} + 4

in the standard form:

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

Therefore parabola properties are:

(h, k) = (0, 4), 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 = 5 x^{2} + 3

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

2 y = 4 x - x^{2}

y = 2 x^{2} - 4 x - 6

y = 6 x^{2} - 37 x - 35