Find the minimum value of the quadratic equation: y = x^2 + 4x - 12
To find the minimum point of a quadratic equation, the quadratic should be written in the form '(x+a)2+ b', i.e. completing the square, where the coordinates of the minimum point will be (-a,b). To complete the square of the above equation, halve the coefficient of x (number before x) to find the value of 'a' that goes inside the bracket - this is 4 divided by 2, which is 2. To find the letter 'b', take the constant from the end of the quadratic (in this case -12) and subtract the value of a2 from it, so b = -12 - (2*2) = -12 - 4 = -16. Combining these, the quadratic can be written in the completed square form: y = (x+2)2-16. This answer can be checked by multiplying out the brackets and seeing if it is equal to the original quadratic: (x+2)2-16 = x2+ 2x + 2x + 4 - 16 = x2 + 4x = 12 --> this is equal to the original quadratic equation, to the completed square form y = (x+2)2-16 is correct, giving a minimum point of (-2, -16).
