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

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axis y=5x^2+x-4

axis

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

x = - \frac{1}{10}

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{20} (y - (- \frac{81}{20})) = (x - (- \frac{1}{10}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{10}, - \frac{81}{20}), 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}{10}

y = + \frac{17}{4} (x + 2)^{2}

y = + 2 x^{2} + x - 1

x = (y - 4)^{2} + 3

y = 2.1 + 4.4 x - 0.7 x^{2}

y = 2 x^{2} + 15 x