What is the axis y=(1+x^2-8x)/5 ?

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axis y=(1+x^2-8x)/5

axis

y = \frac{1 + x ^{2} - 8 x}{5}

x = 4

Solution steps

y = \frac{1 + x ^{2} - 8 x}{5}

Rewrite

y = \frac{1 + x ^{2} - 8 x}{5}

in the standard form

4 \cdot \frac{5}{4} (y - (- 3)) = (x - 4)^{2}

Therefore parabola properties are:

(h, k) = (4, - 3), p = \frac{5}{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 = 4

y^{2} + 6 x - 4 y - 11 = 0

(x + 3)^{2} = 2 (y - 4)

+ 3 y^{2} = 24 x

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

y = - \frac{13}{11} (x + 3)^{2} + 1