I need to find the vertex and axis of symmetry of the folowing quadratic function. Can someone PLEASE help me!!
y= -x^2-6x
You can complete the square ....
y = -(x^2 + 6x +9 - 9
= -( (x+3)^2 - 9)
= -(x+3)^2 + 9
vertex is (-3, 9)
alternate method:
for y = ax^2 + bx + c
the x of the vertex is -b/2a
for for yours
the x of the vertex is -(-6)/-2 = -3
sub into the function ...
y = -9 + 18 = 9
so the vertex is (-3,9) as above
The simplest way is to use Calculus, but I am not sure if you are studying Calculus.
Reiny,
It seems like Calculus, but it actually falls under college algebra. Because I can get some calculus tutoring and the answers fit. Thank you for the help, it is greatlt appreciated!!!
Of course, I'd be happy to help you find the vertex and axis of symmetry of the quadratic function `y = -x^2 - 6x`.
To find the vertex of a quadratic function in the form `y = ax^2 + bx + c`, you can use the formula `x = -b/2a` to find the x-coordinate of the vertex, and then substitute it back into the function to find the corresponding y-coordinate.
In this case, the function `y = -x^2 - 6x` is already in the standard form, with `a = -1`, `b = -6`, and `c = 0`. To find the x-coordinate of the vertex, use the formula:
`x = -b / (2a)`
`x = -(-6) / (2 * (-1))`
`x = 6 / (-2)`
`x = -3`
Now, substitute `x = -3` back into the function to find the y-coordinate:
`y = -(-3)^2 - 6(-3)`
`y = -9 + 18`
`y = 9`
The vertex of the quadratic function is `(x, y) = (-3, 9)`. The x-coordinate of the vertex is -3, and the y-coordinate is 9.
To find the axis of symmetry, you simply take the x-coordinate of the vertex. Therefore, the axis of symmetry for the given quadratic function is `x = -3`.
So, the vertex is (-3, 9) and the axis of symmetry is x = -3.