Asked by ladybug
I need to find the vertex and axis of symmetry of the folowing quadratic function. Can someone PLEASE help me!!
y= -x^2-6x
y= -x^2-6x
Answers
Answered by
Reiny
You can complete the square ....
y = -(x^2 + 6x <b>+9 - 9</b>
= -( (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.
y = -(x^2 + 6x <b>+9 - 9</b>
= -( (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.
Answered by
ladybug
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!!!
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!!!
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