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

The axis y<= 2x^2-10x+10 is x= 5/2

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axis y<= 2x^2-10x+10

axis

y \leq 2 x^{2} - 10 x + 10

x = \frac{5}{2}

Solution steps

y \leq 2 x^{2} - 10 x + 10

Rewrite

y \leq 2 x^{2} - 10 x + 10

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{5}{2})) = (x - \frac{5}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{5}{2}, - \frac{5}{2}), p = \frac{1}{8}

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{5}{2}

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

y = (x - \frac{24}{x + 2}) (x - \frac{12}{x - 4})

x^{2} = - 8 y

y^{2} + 4 x + 1 = 0

y = - \frac{1}{6} x^{2} + 4 x + 10