Vertex of a Parabola?

Question

How do I find the vertex of the parabola
f(x)= (-x)^2 +9x -5

Answer 1

y=x^2+9x-5 =(x+4.5)2 -25.25
so the vertex is the point (-4.5,-25.25)

Answer 2

lets pretend that the equation you're dealing with is y=x^2-7x+12.
the first thing that you have to do is find the line of symmetry. to do this, you substitute the a and b numbers into the equation:
x=-(b/2(a)) this would mean that you have x=-(-7/2(1)) this would end up being x=7/2. This is your line of symmetry, which your vertex will be on.
You then substitute 7/2 into the original equation where you see x. this would make the equation y=49/4-98/4+48/4 this then simplifies to y=-1/4. Your vertex is at (7/2,-1/4)

With your problem, you do the same thing, but you have to substitute the value of x in where you see f(X)

Answer 3

Easy way to remember this:
If the parabola is given by y = ax² + bx + c,
the x coordinate of the vertex is always -b/2a.
Your equation is the same as
y = x²+9x-5.
So in your problem the x coordinate of the vertex is
x = -9/2 = -4.5
Now get the y coordinate by substituting this
value into the equation.
y = (-4.5)² - 9(4.5) -5 = 20.25 - 40.5 -5
= -25.25.
The vertex is at (-4.5, -25.25).

Answer 4

partial derivate with respect to x the given equation
so,2x+9=0 therefore x-coordinate of vertex is -9/2 and y coordinate is 0
vertex of parabola=(-9/2,0)
hope m clear..................