What is the turning point of a quadratic equation y=4^2+8x+3?
It can also be called a “critical point” If you are using derivatives, then you set the derivative equal to 0. If, like me, you feel too lazy for that, then you find the roots. The critical point is always half way between the roots, because the quadratic is a symmetrical curve (the left side is a mirror image of the right side). Imagine the equation written as y = ax^2 + bx + c We know the roots (if they exist) are at x = [ -b +/- sqrt( b^2 – 4ac) ] / 2a distribute the denominator: x = -b/2a +/- [sqrt(b^2 – 4ac)]/2a This gives us two roots (call them x1 and x2); to find the half-way point, we add them together, then divide by 2. (x1 + x2)/2 = half of {-b/2a + [sqrt(b^2 – 4ac)]/2a} + { -b/2a – [sqrt(b^2 – 4ac)]/2a} The sqrt portion cancels out (one is + and the other is -) leaving you with [(-b/2a) + (-b/2a)]/2 = -b/2a In your equation, we have a = +4 b = +8 c = +3 (we don’t need it for the critical point) critical point = -b/2a = -8/2(4) = -8/8 = -1 The x-value of the critical point i
