What is the axis y=5x^2-9 ?

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axis y=5x^2-9

axis

y = 5 x^{2} - 9

x = 0

Solution steps

y = 5 x^{2} - 9

Rewrite

y = 5 x^{2} - 9

in the standard form:

4 \cdot \frac{1}{20} (y - (- 9)) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, - 9), 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

4 x^{2} - 24 x - 40 y - 4 = 0

x = + (y + 11)^{2} - 14

y = x^{2} - 3 x + 28

y = (x + 1) x

y = \frac{2 - x ^{2} + 2 x}{3}