What is the axis y=2x^2+12x+10 ?

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axis y=2x^2+12x+10

axis

y = 2 x^{2} + 12 x + 10

x = - 3

Solution steps

y = 2 x^{2} + 12 x + 10

Rewrite

y = 2 x^{2} + 12 x + 10

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 3, - 8), p = \frac{1}{8}

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 = - 3

y = 290 (8 x + 0.5 x^{2} - 1.4)

y = 6 x^{2} + 6 x - 36

y = 5 x^{2} - 4 x + 3

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

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