What is the axis y=-6(x-5)^2+12 ?

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

axis y=-6(x-5)^2+12

axis

y = - 6 (x - 5)^{2} + 12

x = 5

Solution steps

y = - 6 (x - 5)^{2} + 12

Rewrite

y = - 6 (x - 5)^{2} + 12

in the standard form:

4 (- \frac{1}{24}) (y - 12) = (x - 5)^{2}

Therefore parabola properties are:

(h, k) = (5, 12), p = - \frac{1}{24}

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 = - 2 x^{2} + x - 1

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

y = x^{2} + 14 x + 48

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

(y - 7)^{2} = - 12 (x + 1)