nice attempt :) at least you couldve said some stupid crips rolling in 6 fo's with dicks or something, which ive herd before
the answer is -6 -apex-
There are three main types of vertices for an absolute value function. There are some vertices which are carried over from the function, and taking its absolute value makes no difference. For example, the vertex of the parabola y = 3*x^2 + 15 is not affected by taking absolute values. Then there are some vertices which are reflected in the x-axis because of the absolute value. For example, the vertex of the absolute value of y = 3*x^2 - 15, that is y = |3*x^2 - 15| will be the reflection of the vertex of the original. Finally there are points where the function is "bounced" off the x-axis. These points can be identified by solving for the roots of the original equation. -------------- The above answer considers the absolute value of a parabola. There is a simpler, more common function, y = lxl. In this form, the vertex is (0,0). A more general form is y = lx-hl +k, where y = lxl has been translated h units to the right and k units up. This function has its vertex at (h,k). Finally, for y = albx-hl + k, where the graph has been stretched vertically by a factor of a and compressed horizontally by a factor of b, the vertex will be at (h/b,ak). Of course, you can always find the vertex by graphing, especially since you might not remember the 2nd or 3rd parts above.
By completing the square y = (x+3)2+1 Axis of symmetry and vertex: x = -3 and (-3, 1) Note that the parabola has no x intercepts because the discriminant is less than zero
To find the extreme value of the parabola y = x2 - 4x + 3 ...(1) Take the derivative of the equation.y = x2 - 4x + 3y' = 2x - 4(2) Set the derivative = 0 and solve for x.y' = 2x - 40 = 2x - 42x = 4x = 4/2x = 2(3) Plug this x value back into the original equation to find the associated y coordinate.x = 2y = x2 - 4x + 3y = (2)2 - 4(2) + 3y = 4 - 8 + 3y = -1So the vertex is at (2, -1).
The only way I ever learned to find it was to think about it. The function f(x) = log(x) only exists of 'x' is positive. As 'x' gets smaller and smaller, the function asymptotically approaches the y-axis.
Find the measure of this angles m1 equals 123 m8 equals?
You would convert it to vertex form by completing the square. You can also find the optimum value as optimum value and vertex are the same.
-2-5
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
The given equation is y = x - 4x + 2 which can be written as y = -3x + 2 This is an equation of a straight line. Therefore it has no vertex and so cannot be written in vertex form.
The vertex angle is connected to the vertex point
y = x2 + 12x + 21At the max or min point, the first derivative of the function = 0.2x + 12 = 02x = -12x = -6
jaj no se kompas jaj
You can find a vertex wherever two lines (or line segments) meet.
25.12
12.56
It depends on the vertex of what!
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)