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

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axis y=5x^2+2x-4

axis

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

x = - \frac{1}{5}

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{20} (y - (- \frac{21}{5})) = (x - (- \frac{1}{5}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{5}, - \frac{21}{5}), 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 = - \frac{1}{5}

y = 2 x^{2} - x - 10

y = 3 x^{2} + 2 x + 6

x^{2} = - \frac{4}{5} (y - 2)

5 x - 6 - \frac{1}{2} y^{2} + 2 y^{2} = 0

y = x^{2} + 12 x + 6