Start by finding the x-coordinate of that vertex. You can do that by taking the derivative of the function and solving for zero:
y = -4x2 - 16x - 11
y' = -8x - 16
0 = -8x - 16
8x = -16
x = -2
Now simply plug the x-coordinate into the original equation to get your y-coordinate:
y = -4(-2)2 - 16(-2) - 11
y = -16 + 32 - 11
y = 5
So the vertex occurs at the point (-2, 5)
Alternative answer:
It should be noted that the above method requires a little extra work if you are not working with a parabola, as the first derivative only allows you to find the critical points of a function. With an arbitrary function, you also need to take the second derivative of the function to determine if the critical point is a maximum, minimum or point of inflection.
If you don't know Calculus yet, there is also an algebraic method to find the vertex of a parabola
We have:
y = -4x^2 - 16x - 11
This is the parabola's standard form. In order to find the vertex through algebra we need to convert it to vertex form:
y = a(x - h) + k
Where (h,k) is our vertex
We can do this by factoring (which is always a pain):
First factor out a -4:
y = -4(x^2 + 4x) - 11
We left the -11 alone because 11/4 is pretty ugly to work with, so we will leave it on the side for now.
Next we complete the square:
y = -4(x^2 + 4x + c - c) - 11
y = -4(x^2 + 4x + c) - 11 + 4c
We want to find c, such that 2*(c^(1/2)) = 4, which gives us c = 4.
y = -4(x^2 + 4x + 4) - 11 + 16
y = -4(x + 2)^2 + 5
Comparing with our vertex form:
y = a(x - h) + k
We have a = -4, h = -2 and k = 5.
(h, k) is our vertex, which gives us (-2, 5). This is consistent with the answer given by the Calculus method above, which is reassuring.
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f(x) = -4x2 - 16x - 11 a = -4, b = -16, c = -11 x-coordinate of the vertex = -b/2a = -(-16)/2(-4) = 16/-8 = -2 y-coordinate = f(-2) = -4(-2)2 -16(-2) - 11 = -16 + 32 - 11 = 5 vertex is (-2, 5)
4x2 - 16x + 12
If your function is 4x^2+16x+16=0 then x=-2, if it is 4x^2-16x+16=0, then x=2
I think you are talking about the x-intercepts. You can find the zeros of the equation of the parabola y=ax2 +bx+c by setting y equal to 0 and finding the corresponding x values. These will be the "roots" of the parabola.
It is; (2x+4)^2.