What is the axis (2x-7)^2+8 ?

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axis (2x-7)^2+8

axis

(2 x - 7)^{2} + 8

x = \frac{7}{2}

Solution steps

y = (2 x - 7)^{2} + 8

Rewrite

y = (2 x - 7)^{2} + 8

in the standard form:

4 \cdot \frac{1}{16} (y - 8) = (x - \frac{7}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{7}{2}, 8), 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{7}{2}

x = 2 y^{2} - 1 - (5 y - 6)

y = \frac{x ( 3 x - 1 )}{2}

y = + 2 (x + 1)^{2} - 3

y = \frac{x ^{2} - 4 x}{11}

\frac{a + 1}{b} = \frac{a - 1}{b} + \frac{b + 1}{a}