Algebra: A standard parabola is y = x^2. Its vertex is at (0,0) You can change the position (or vertex) of the parabola. To move a parabola across the x-axis, you can add or subtract a number from x WITHIN brackets of the ^2 eg. (x + 1)^2 will move the parabola across the x-axis. It will move is one unit to the LEFT (as the sign is opposite to the direction it moves ie. The sign it + but you move the whole parabola in the -ve direction). Adding or subtracting a number from x OUTSIDE of the ^2 moves the parabola up or down the y-axis eg. x^2 + 3 will move the parabola UP 3 units (the sign is the same as the direction it moves when the added/subtracted number is outside of the ^2 ie. the sign is positive so the parabola moves up in the positive direction)
From this, we can conclude that because (x + 1)^2 + 3, the vertex will be where x = -1 and where y = 3
Vertex : (-1,3)
Calculus: f(x) = (x + 1)^2 + 3 = x^2 + 2x + 1 + 3 = x^2 + 2x + 4 Expanding the formular to make it easier to differentiate
f'(x) = 2x + 2 Differentiating (finding the formular the the gradiet of the parabola)
0 = 2x + 2 When the gradient is equal to zero, it must be the vertex
-2 = 2x -2/2 = x x = -1 Solve to give the x value at the vertex
f(x) = (x + 1)^2 + 3 = (-1 + 1)^2 + 3 Substitute x = -1 into original equatiom to find y value at the vertex
