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

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

axis

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

x = - 1

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

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

y = \frac{4 x ^{3} - 7 x ^{2} - 11 x + 5}{4 x + 5}

x^{2} - 2 x - 4 y + 17 = 0

y = 2 x^{2} - 3 x

y = 3 x^{2} - 6 x - 24

y = 4 x^{2} + 8 x + 13