What is the axis y=x^2-8x+15 ?

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

axis y=x^2-8x+15

axis

y = x^{2} - 8 x + 15

x = 4

Solution steps

y = x^{2} - 8 x + 15

Rewrite

y = x^{2} - 8 x + 15

in the standard form

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

Therefore parabola properties are:

(h, k) = (4, - 1), 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 = 4

y^{2} = 16 x

4 x^{2} - 20 x - 24 y + 97 = 0

y = x^{2} + 3 x - 4

y = 2 x^{2} - 2

y = x^{2} - 13 x + 40