The slope of the tangent at any point P (U, V) of the parabola y = x ^ 2 - 3x + 2 is calculated, and the equation of the tangent at the vertex of the parabola is obtained The answer I got is y = - 1 / 4

The slope of the tangent at any point P (U, V) of the parabola y = x ^ 2 - 3x + 2 is calculated, and the equation of the tangent at the vertex of the parabola is obtained The answer I got is y = - 1 / 4

Finding the derivative dy / DX = 2x-3
The slope of the tangent at P (U, V) is 2u-3
Let 2x-3 = 0 get x = 3 / 2, that is, the abscissa of the vertex, and substitute it into the original equation to get the ordinate y = - 1 / 4,
Vertex coordinates are (3 / 2, - 1 / 4)
Because the quadratic coefficient a = 1 is greater than zero, the slope at the vertex is zero with the opening upward, and the equation is
y-(-1/4)=0(x-3/2)
That is y = - 1 / 4