What is the axis y=+x^2+3x+14 ?

The axis y=+x^2+3x+14 is x=-3/2

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axis y=+x^2+3x+14

axis

y = + x^{2} + 3 x + 14

x = - \frac{3}{2}

Solution steps

y = x^{2} + 3 x + 14

Rewrite

y = x^{2} + 3 x + 14

in the standard form

4 \cdot \frac{1}{4} (y - \frac{47}{4}) = (x - (- \frac{3}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{2}, \frac{47}{4}), p = \frac{1}{4}

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}{2}

4 x^{2} - 12 x - 16 y + 41 = 0

x = (y - 12)^{2} - 156

x = \frac{1}{4} y^{2}

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

\frac{y - 2}{4} - \frac{x ^{2}}{9} = 1