What is the axis y=-63/16+1/16 x^2-1/8 x ?

The axis y=-63/16+1/16 x^2-1/8 x is x=1

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axis y=-63/16+1/16 x^2-1/8 x

axis

y = - \frac{63}{16} + \frac{1}{16} x^{2} - \frac{1}{8} x

x = 1

Solution steps

y = - \frac{63}{16} + \frac{1}{16} x^{2} - \frac{1}{8} x

Rewrite

y = - \frac{63}{16} + \frac{1}{16} x^{2} - \frac{1}{8} x

in the standard form

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

Therefore parabola properties are:

(h, k) = (1, - 4), p = 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 = x^{2} + 8 x - 5

y = 3 x^{2} - 5 x 7

y = 7 x^{2} - 10 x + 5

y = - \frac{1}{2} x^{2} - 4 x + 10

y = x^{2} - x - 20

What is the axis y=-63/16+1/16 x^2-1/8 x ?
The axis y=-63/16+1/16 x^2-1/8 x is x=1