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

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axis y=(2x+1)^2

axis

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

x = - \frac{1}{2}

Solution steps

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

Rewrite

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

in the standard form:

4 \cdot \frac{1}{16} (y - 0) = (x - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{2}, 0), p = \frac{1}{16}

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 = - \frac{1}{2}

y = 12 x^{2}

y^{2} = 28 x

y = 0.00225 x^{2} + 5999.9325 x + 0.50625

y = \frac{4 x ^{2} + 12 x - 15}{3}

(x + 2)^{2} = - 12 (y - 5)