Finding the tangent equation of parabola y = x2 power at x = 2

Finding the tangent equation of parabola y = x2 power at x = 2

Parabola y = x ^ 2
Derivative y '= 2x
So the slope of the tangent of the parabola y = x ^ 2 at x = 2 is k = 2 * 2 = 4
And when x = 2, y = 2 ^ 2 = 4
So the tangent equation is y-4 = 4 (X-2)
That is y = 4x-4

It is known that the parabola y = ax + BX + C passes through points a (5,0), B (6, - 6) and the origin
1. Find the function relation of parabola

）Put in three
25a+5b+c=0
36a+6b+c=-6
c=0
The solution is a = - 1, B = 5, C = 0
So the function relation of parabola is y = - x ^ 2 + 5x

Given the image of parabola y = ax ^ 2 + BX + C, then the equation AX ^ 2 + BX + C-3 of X is rooted
The image can't be uploaded. The opening is facing down, the symmetry axis is on the right side of Y, intersects the Y axis and the positive half axis, and the vertical coordinate of the vertex is 3

The ordinate of the vertex is 3,
therefore
Original formula - 3 = translation down 3 unit length
The vertex ordinate becomes zero
There is only one intersection between the image and the x-axis
So the original equation has two identical real solutions