Find the vertex of the parabola y -4x2 - 16x - 11? - Answers

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.