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What is the vertex of the parabola y-2x2-4x 4?
Assuming the missing symbol there is an equals sign, then we have: y - 2x2 - 4x = 4 We can find it's vertex very easily by solving for y, and finding where it's derivative equals zero: y = 2x2 + 4x + 4 y' = 4x + 4 0 = 4x + 4 x = -1 So the vertex occurs Where x = -1. Now we can plug that back into the original equation to find y: y = 2x2 + 4x + 4 y = 2 - 4 + 4 y = 2 So the vertex is at the point (-1, 2)
Find the vertex of y equals 3x2-4x equals 3?
y=3x^2-4x=3 y=3x^2-4x-3 y = ax^2 + bx + c is the parabola y = 3x^2-4x-3 x-coordinate of vertex is -b/2a = -(-4) / (2)(3) = +4 / (2)(3) = 4/6=2/3 y coordinate of the vertex: y = ax^2+bx+c y = 3(2/3)^2-4(2/3)-3 y= 3(4/9) -8/3 -3 y=12/9-8/3-3 = -13/3 Vertex : ( 2/3, , -13/3 )
What is the relationship between a function and its derivative?
The derivative if a function is basically it's slope, or its rate of change. An example is the function y = 4x - 6. This is a line with a slope of 4. The derivative is y' = 4. Another example is the function y = 3x2. This is a parabola with a vertex at (0,0). Its derivative is y' = 6x. At x = 0, the slope of the parabola is 6*0, which is 0, since this is the vertex of the parabola. To the left, at x is -4 for example, the derivative (and therefore slope) is negative. To the right, at x = 5 for example, the derivative is positive. The farther away from the vertex, the greater the value of the derivative so the the slope of the function increases as you move away from the vertex (it gets steeper).
Find the vertex of the parabola y -4x2 - 16x - 11?
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.
Solve 4x equals 72?
4x = 72.divide both sides by 4x = 18
