What is the axis y=3x^2+6x+10 ?

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axis y=3x^2+6x+10

axis

y = 3 x^{2} + 6 x + 10

x = - 1

Solution steps

y = 3 x^{2} + 6 x + 10

Rewrite

y = 3 x^{2} + 6 x + 10

in the standard form

4 \cdot \frac{1}{12} (y - 7) = (x - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (- 1, 7), p = \frac{1}{12}

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 = - 1

F (x) = 2 x^{2} + 2 x + 3

y = 10 (x - 6)^{2} + 5

y = 11 x - x^{2}

y = - \frac{1}{2} (x + 6)^{2} + 8

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