What is the axis y=4x^2+3x-1 ?

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axis y=4x^2+3x-1

axis

y = 4 x^{2} + 3 x - 1

x = - \frac{3}{8}

Solution steps

y = 4 x^{2} + 3 x - 1

Rewrite

y = 4 x^{2} + 3 x - 1

in the standard form

4 \cdot \frac{1}{16} (y - (- \frac{25}{16})) = (x - (- \frac{3}{8}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{8}, - \frac{25}{16}), 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 = - \frac{3}{8}

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

y = 2 x^{2} - 3 x - 5

y = 2 x^{2} - 3 x - 8

y = 6 x^{2} + 5 x + 3

y = 6 x^{2} + 5 x + 7