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

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axis y=7x^2+5

axis

y = 7 x^{2} + 5

x = 0

Solution steps

y = 7 x^{2} + 5

Rewrite

y = 7 x^{2} + 5

in the standard form:

4 \cdot \frac{1}{28} (y - 5) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 5), 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} + 8 x

y = - 16 x^{2} + 40 x + 2

x = - 2 y + y (y - 2) - 2 y (y - 1) - (y - 3)

y = - 0.5 x^{2} + 5.5

y = \frac{1}{2} x^{2} + 9 x + 8