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First you have to identify the values of a, b, and c of the given quadratic equation in the general form:
    ax² + bx + c = 0

vertex is (h, k) 

  h = -b/2a

  k = 4ac -b²

h is x coordinate, and k is the y-coordinate of the vertex (h, k).

Note that if the parabola (graph of the quadratic equation) opens downward (inverted U), the vertex is the maximum.  While if the parabola opens upward (U-shaped), the vertex is the minimum. 

To know without graphing if the parabola opens downward, a is negative; opens upward, a is positive.
0 0 0
is this using differential calculus?
No, i missed the word differentiation. Let me report my answer and I'll post the correct one :-)
you should have knowledge of solving linear equation system by substitution or elimination method.What's I wrote as answer above is the same concept with regards to the nature of the parabola (minimum, maximum. Work from the first derivative of quadratic function which is : f(x) = ax^2 + bx + c, the first derivative is f' ' (x) = 2ax + b.You can obtain two linear equations related to the x coordinates of the two tangent lines, to solve its system (of the tangents). Get a Leithold book,please.
ok thankzs
**what I wrote, not what's I wrote. Typo :-)
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