What is the axis y=8x^2+12x+3 ?

The axis y=8x^2+12x+3 is x=-3/4

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axis y=8x^2+12x+3

axis

y = 8 x^{2} + 12 x + 3

x = - \frac{3}{4}

Solution steps

y = 8 x^{2} + 12 x + 3

Rewrite

y = 8 x^{2} + 12 x + 3

in the standard form

4 \cdot \frac{1}{32} (y - (- \frac{3}{2})) = (x - (- \frac{3}{4}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{4}, - \frac{3}{2}), 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{3}{4}

y = 4 x^{2} - 8 x + 7

x^{2} = - 6 y

+ 5 x^{2} + 30 x + 24 y = 51

x^{2} = - 4 y

y = - 5 x^{2} + 118 x + 7