Write f(x)=x2−6x+16f(x)=x2−6x+16 in the form f(x)=(x−3)2+pf(x)=(x−3)2+p and hence state the range of ff.+ Check answer

To write the function f(x) = x^2 - 6x + 16 in the form f(x) = (x - 3)^2 + p, we need to complete the square.

The general form for completing the square is (x - h)^2 + k, where h is half the coefficient of x and k is the constant term.

Step 1: Identify the coefficient of x, which is -6. Half of -6 is -3.

Step 2: Substitute -3 into the equation to get (x - 3)^2.

Step 3: Expand (x - 3)^2 to get x^2 - 6x + 9.

Step 4: Compare the expanded form x^2 - 6x + 9 with the original equation x^2 - 6x + 16. We can see that the constant term in the original equation is 16, while in the expanded form it is 9.

Step 5: To make these two equations equal, we need to add a constant p to the expanded form. The constant p is the difference between the constant term in the original equation and the constant term in the expanded form, which is 16 - 9 = 7.

So, f(x) = (x - 3)^2 + 7.

The range of a quadratic function in the form f(x) = (x - h)^2 + k is [k, ∞) if the coefficient of (x - h)^2 is positive, and (-∞, k] if the coefficient of (x - h)^2 is negative.

Since the coefficient of (x - 3)^2 in f(x) = (x - 3)^2 + 7 is positive, the range of f is [7, ∞).