What is the axis y=x^2+10x+21 ?

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axis y=x^2+10x+21

axis

y = x^{2} + 10 x + 21

x = - 5

Solution steps

y = x^{2} + 10 x + 21

Rewrite

y = x^{2} + 10 x + 21

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 5, - 4), 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

y = 5 x^{2} - 12 x + 7

x^{2} - 6 x - 4 y = - 1

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

y = 4 - x (x + 3)

y^{2} = 16 - 4 x