What is the axis y=x^2-10x-7 ?

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axis y=x^2-10x-7

axis

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

x = 5

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{4} (y - (- 32)) = (x - 5)^{2}

Therefore parabola properties are:

(h, k) = (5, - 32), p = \frac{1}{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 = 5

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

y = x^{2} - 6 x + 18

x + 5 + (y - 3)^{2} = 0

y = - x^{2} + 1

y = 4 x^{2} - 40 x + 98