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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

axis

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

x = 3

Solution steps

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

Rewrite

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

in the standard form

4 (- \frac{1}{4}) (y - 16) = (x - 3)^{2}

Therefore parabola properties are:

(h, k) = (3, 16), 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 = 3

12 (x - 2) = (y + 2)^{2}

(y - 2)^{2} = 4 (- 10) (x - 2)

y^{2} = - 2 x

y = x^{2} - 3 x - 1

x^{2} - 2 x + 36 y + 28 = 0