How do you find the vertex of a parabola and an absolute-value graph with just the equation?

Answer 1

For a parabola ax^2 + bx + c the x coordinate of the vertex is
-b/2a and y can be worked at as f(-b/2a)

For any equation to find turning points (vertices) you differentiate as the gradiant must be zero. To decide if the vertex is a local maximum (peak) or minimum (trough) or a point of inflexion you have to differentiate again. If the second derivative is +ve you have a local minimum value (trough), if the derivative is -ve you have a local maximum (peak), if you find the derivative is zero then you have a pointof inflexion.

FYI - if you don't know already then the derivative of ax^n is anx^(n-1) for all n

e.g. y= x^2 - 10x + 6

we use dy/dx to represent the dervative of y with respect to x and d2y/dx2 to represent the derivative of that derivative

dy/dx = 2x -10. This equals zero when x = 5 so the vertex is at the point when x = 5 (-19)
d2y/dx2 = 2 +ve so this is a minimum (trough) the parabola is U shaped.

Answer 2

If the equation of the parabols is in the form y = ax^2 + bx + c = 0, then the x-coordinate of the vertex will be x = -b/2a
Plug this into f(x) and you'll get the y-coordinate of the vertex.
For the absolute value, just set the inside of the absolute value equation to 0 and that will give you the x-coordinate of the vertex. Plug this x into the function for the y-coordinate.

Answer 3

Get the equation into the form y = (x - h)^2 + k. The parabola vertex is then at (h,k)

Answer 4

Find the derivative.