What is the axis y=2(x+3)^2+4 ?

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axis y=2(x+3)^2+4

axis

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

x = - 3

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

(h, k) = (- 3, 4), 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

f (x) = - 3 x^{2} + 5 x - 2

- x^{2} + 10 x - 24

y = \frac{x ^{2} - 16 x + 70}{2}

x = (2 y + 5) (y + 6)

y = x^{2} + 12 x + 43