What is the axis y=10x^2+1-6x ?

The axis y=10x^2+1-6x is x= 3/10

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axis y=10x^2+1-6x

axis

y = 10 x^{2} + 1 - 6 x

x = \frac{3}{10}

Solution steps

y = 10 x^{2} + 1 - 6 x

Rewrite

y = 10 x^{2} + 1 - 6 x

in the standard form

4 \cdot \frac{1}{40} (y - \frac{1}{10}) = (x - \frac{3}{10})^{2}

Therefore parabola properties are:

(h, k) = (\frac{3}{10}, \frac{1}{10}), p = \frac{1}{40}

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{3}{10}

y = - 8 x^{2} + 4 x

- 3 y = x^{2} - 8 x + 37

y = - 2 x^{2} + 12 x - 10

y = 2 x^{2} - 12 x + 18

x^{2} - 8 y = 0