What is the axis y=4x^2-16x+11 ?

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axis y=4x^2-16x+11

axis

y = 4 x^{2} - 16 x + 11

x = 2

Solution steps

y = 4 x^{2} - 16 x + 11

Rewrite

y = 4 x^{2} - 16 x + 11

in the standard form

4 \cdot \frac{1}{16} (y - (- 5)) = (x - 2)^{2}

Therefore parabola properties are:

(h, k) = (2, - 5), 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 = 2

y = 3 x - 3 x^{2}

y = + (x - 3)^{2} + 24

y = - x^{2} + 5 x - 2

y = x^{2} - 4

y = x^{2} - 20 x + 100