Question

The vertex form of the quadratic function f(x)= x^2-6x+5 is f(x) = a(x-h)^2+k, what is a, b, and c?

Answer 1

The vertex form of the quadratic function f(x) = x^2 - 6x + 5 is f(x) = 1(x - 3)^2 - 4, where a = 1, h = 3, and k = -4. To find the vertex form, we complete the square by adding and subtracting (b/2)^2 inside the function. The values of 'h' and 'k' give the coordinates of the vertex.

Step-by-step explanation:

The student has provided the quadratic function f(x) = x^2 - 6x + 5 and wants to know its vertex form, which is given by f(x) = a(x-h)^2 + k. To convert the standard form of a quadratic function into vertex form, we complete the square. First, we take the coefficient of the x-term, divide it by 2, and then square it to find the value needed to complete the square.

Let's break this down step-by-step:

1. Given the quadratic function f(x) = x^2 - 6x + 5, we identify a = 1, b = -6, and c = 5 from the standard form ax^2 + bx + c.
2. We take b/2, which is (-6)/2 = -3, and square it to get 9.
3. Add and subtract this value inside the equation to complete the square: f(x) = (x^2 - 6x + 9) - 9 + 5.
4. The expression in the parenthesis is now a perfect square: f(x) = (x - 3)^2 - 4.
5. Thus, the vertex form of the function is f(x) = 1(x - 3)^2 - 4, where a = 1, h = 3, and k = -4.

In the vertex form of a quadratic function, the value of 'a' remains the same as in the standard form, the value of 'h' is the x-coordinate of the vertex, and the value of 'k' is the y-coordinate of the vertex.