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Can you predict the maximum or minimum point on a parabola? - Answers

A parabola is (mathematically speaking) a quadratic function, which looks like this y = ax2 + bx + c where a, b and c are constants. (If three points on the curve are known, then a, b and c can be found.) The gradient, then, can be found by differentiation: dy/dx = 2ax + b A parabola has one maximal or minimal point, where the gradient is zero. 2ax + b = 0 x = -b/2a Use the original function to find the corresponding value of y: y = a(-b/2a)2 + b(-b/2a) + c = b2/4a - b2/2a + c = c - b2/4a So the coordinates of your turning point are ( -b/2a , c - b2/4a ) This result can also be derived by completing the square.



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Can you predict the maximum or minimum point on a parabola? - Answers

https://math.answers.com/other-math/Can_you_predict_the_maximum_or_minimum_point_on_a_parabola

A parabola is (mathematically speaking) a quadratic function, which looks like this y = ax2 + bx + c where a, b and c are constants. (If three points on the curve are known, then a, b and c can be found.) The gradient, then, can be found by differentiation: dy/dx = 2ax + b A parabola has one maximal or minimal point, where the gradient is zero. 2ax + b = 0 x = -b/2a Use the original function to find the corresponding value of y: y = a(-b/2a)2 + b(-b/2a) + c = b2/4a - b2/2a + c = c - b2/4a So the coordinates of your turning point are ( -b/2a , c - b2/4a ) This result can also be derived by completing the square.



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https://math.answers.com/other-math/Can_you_predict_the_maximum_or_minimum_point_on_a_parabola

Can you predict the maximum or minimum point on a parabola? - Answers

A parabola is (mathematically speaking) a quadratic function, which looks like this y = ax2 + bx + c where a, b and c are constants. (If three points on the curve are known, then a, b and c can be found.) The gradient, then, can be found by differentiation: dy/dx = 2ax + b A parabola has one maximal or minimal point, where the gradient is zero. 2ax + b = 0 x = -b/2a Use the original function to find the corresponding value of y: y = a(-b/2a)2 + b(-b/2a) + c = b2/4a - b2/2a + c = c - b2/4a So the coordinates of your turning point are ( -b/2a , c - b2/4a ) This result can also be derived by completing the square.

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      A parabola is (mathematically speaking) a quadratic function, which looks like this y = ax2 + bx + c where a, b and c are constants. (If three points on the curve are known, then a, b and c can be found.) The gradient, then, can be found by differentiation: dy/dx = 2ax + b A parabola has one maximal or minimal point, where the gradient is zero. 2ax + b = 0 x = -b/2a Use the original function to find the corresponding value of y: y = a(-b/2a)2 + b(-b/2a) + c = b2/4a - b2/2a + c = c - b2/4a So the coordinates of your turning point are ( -b/2a , c - b2/4a ) This result can also be derived by completing the square.
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