What is the axis y=+4x^2+7 ?

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axis y=+4x^2+7

axis

y = + 4 x^{2} + 7

x = 0

Solution steps

y = 4 x^{2} + 7

Rewrite

y = 4 x^{2} + 7

in the standard form:

4 \cdot \frac{1}{16} (y - 7) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 7), p = \frac{1}{16}

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} + 5 x + 6

y = x^{2} + 5 x + 4

y = x^{2} + 7 x - 30

y = x^{2} + 8 x + 16

y = - 5 x^{2} - 40 x - 77