What is the vertex of the graph of f (x)

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 = - 4 To make a perfect square we find (b/2) 2. (-4/2) 2 = 4 We 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 = 9/4 We 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).