What is the axis y=(x+4)^2+1 ?

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axis y=(x+4)^2+1

axis

y = (x + 4)^{2} + 1

x = - 4

Solution steps

y = (x + 4)^{2} + 1

Rewrite

y = (x + 4)^{2} + 1

in the standard form:

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

Therefore parabola properties are:

(h, k) = (- 4, 1), 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 = - 4

y^{2} - 12 x + 20 = 0

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

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

(y + 9)^{2} = - 44 (x - 1)

y = 5 x^{2} + 4