What is the axis y=+17/4 (x+2)^2 ?

The axis y=+17/4 (x+2)^2 is x=-2

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axis y=+17/4 (x+2)^2

axis

y = + \frac{17}{4} (x + 2)^{2}

x = - 2

Solution steps

y = \frac{17}{4} (x + 2)^{2}

Rewrite

y = \frac{17}{4} (x + 2)^{2}

in the standard form:

4 \cdot \frac{1}{17} (y - 0) = (x - (- 2))^{2}

Therefore parabola properties are:

(h, k) = (- 2, 0), p = \frac{1}{17}

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 = - 2

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

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

y = 2.1 + 4.4 x - 0.7 x^{2}

y = 2 x^{2} + 15 x

y = x^{2} - x - 12