Find the vertex of the parabola y=x^2-6x?

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atyourservice's profile pic

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

` y=x^2-6x`

`a=1` `b=-6`

Use the formula `-2/(2a)` to find the x

`x= 6/(2(1)) ` `x= 6/2`     `x=3`

plug in the x into the origninal equation

`y=3^2-6(3)`

`y=9-18`

`y= -9`

vertex is `(3,-9)`

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll recall what are the coordinates of the vertex of the parabola:

V(-b/2a ; -delta/4a)

a,b,c are the coefficients of the quadratic:

y = ax^2 + bx + c

Comparing, we'll identify the coefficients:

a = 1, b = -6 , c = 0

We'll calculate the x coordinate of the vertex:

xV = -b/2a

xV = -(-6)/2

xV = 3

We'll calculate the y coordinate of the vertex:

yV = -delta/4a

delta = b^2 - 4ac

delta = 36 - 4*1*0

delta = 36

yV = -36/4

yV = -9

The requested coordinates of the vertex of the given parabola are: V(3 , -9).

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