What is the axis 6x^2-7x+y=0 ?

The axis 6x^2-7x+y=0 is x= 7/12

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axis 6x^2-7x+y=0

axis

6 x^{2} - 7 x + y = 0

x = \frac{7}{12}

Solution steps

6 x^{2} - 7 x + y = 0

Rewrite

6 x^{2} - 7 x + y = 0

in the standard form

4 (- \frac{1}{24}) (y - \frac{49}{24}) = (x - \frac{7}{12})^{2}

Therefore parabola properties are:

(h, k) = (\frac{7}{12}, \frac{49}{24}), p = - \frac{1}{24}

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}{12}

+ (x - 2) = 3 (y - 5)^{2}

x^{2} - 2 x - 4 y + 13 = 0

(x - 1)^{2} = - 8 (y - 4)

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

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