Here are some more examples:
Let's find the vertex of the parabola given by the function:f(x) = x2 - 4x + 7.To complete the square we look at the middle term's coefficient.b = -4To make a perfect square we find (b/2)2.(-4/2)2 = 4We add 4 to the first two terms and subtract 4 from the constant term.f(x) = x2 - 4x + 4 + 7 - 4 = x2 - 4x + 4 + 3.We rewrite the expression as a perfect square.x2 - 4x + 4 = (x - 2)2 which is (x + b/2)2.We substitute the perfect square into the function.f(x) = (x - 2)2 + 3.According to the standard form of a quadratic function the vertex is (2,3).
Let's find the vertex of the parabola given by the function:f(x) = x2 - 3x + 2.To complete the square we first look at the middle term's coefficient.b = -3.To make a perfect square we find (b/2)2.(-3/2)2 = 9/4We add 9/4 to the first two termsn and subtract 9/4 from the constant term.f(x) = x2 - 3x + 9/4 + 2 - 9/4 = x2 - 3x + 9/4 - 1/4.We rewrite the expression as a perfect square.x2 - 3x + 9/4 = (x - 3/2)2 which is (x + b/2)2.We substitute the perfect square into the function.f(x) = (x - 3/2)2 - 1/4.According to the standard form of a quadratic function the vertex is (3/2,-1/4).
Let's find the vertex of the parabola given by the function:f(x) = 2x2 + 4x - 3.To complete the square we need to first factor out a factor of 2 from the first two terms.f(x) = 2(x2 + 2x) - 3.Next we look at the middle term's coefficient.b = 2.To make a perfect square in the parenthesis we find (b/2)2.(2/2)2 = 1.We add 1 inside the parenthesis, since the expression inside the parenthesis is multiplied by 2, we are really adding 2. We subtract 2 from the constant term.f(x) = 2(x2 + 2x + 1) - 3 - 2 = 2(x2 + 2x + 1) - 5.We rewrite the expression as a perfect square.x2 + 2x + 1 = (x + 1)2 which is (x + b/2)2.We substitute the perfect square into the function.2(x + 1)2 - 5.According to the standard form of a quadratic function the vertex is (-1,-5).