What is the axis y=8x^2+7x-1 ?

The axis y=8x^2+7x-1 is x=-7/16

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axis y=8x^2+7x-1

axis

y = 8 x^{2} + 7 x - 1

x = - \frac{7}{16}

Solution steps

y = 8 x^{2} + 7 x - 1

Rewrite

y = 8 x^{2} + 7 x - 1

in the standard form

4 \cdot \frac{1}{32} (y - (- \frac{81}{32})) = (x - (- \frac{7}{16}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{7}{16}, - \frac{81}{32}), p = \frac{1}{32}

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{7}{16}

y = x^{2} - 64

x = 3 y^{2}

4 y = - x^{2} - 12 x - 48

y = - 4 (x + 10)^{2} + 6

(x + 1)^{2} = - 3 (y - 6)