Question

Find the vertex of the parabola. State whether the vertex is a max or a min.
f(x) = -2x2 + 12x + 13

Answer 1

a = -2 b = 12 c = 13
Substitute it into the formula, then there's your vertex
V = (-b/2a , a(-b/2a)^2 + b(-b/2a) +c)

Answer 2

The vertex of a parabola is the point where the derivative is zero:

\[f'(x) = -4x + 12 = 0\]
\[x = 3\]
that's the x coordinate of our vertex point
to find the y coordinate, we must find f(3) (the value of y at x = 3)
\[f(3) = -18 + 36 + 13 = 31\]
so the vertex of f(x) is at (3,31)
to find out whether it is a minimum or a maximum, we find the second derivative of f(x):\[f''(x) = -4\]since f''(x) is negative, the graph is concave down at that poin. Therefore, the vertex is a maximum.

Answer 3

What ? It says find the vertex not the Point Of Inflexion ?

Answer 4

exactly

Answer 5

So am I wrong ?

Answer 6

actually, your equation does the exact same thing as my working out. You are also right.

Answer 7

Oh, so you took extra steps xD

Answer 8

no extra steps. the only extra thing I did was to find out whether the point was a minimum or a maximum, which you didn't do in your answer

Answer 9

f(x)=-2(x^2-6x)+13
f(x)=-2(x-3)^2+13-(9.(-2))
f(x)=-2(x^2-6x)+13+18
f(x)=-2(x^2-6x)+31
vertex ( 3,31)
maximum , open down
Domain: all real number
\[Range: (-\infty,31)\]

Answer 10

ty

Answer 11

Yeah, I didn't do the minimum or maximum because I just showed how to do it xD . The person can find the minimum or maximum easily anyway xD