Determine the vertex of the quadratic function f(x)=x^2 - 10x +24 and rewrite it in vertex form

Answer 1

Answer 2

When the leading coefficient of a quadratic function is 1, completing the square to transform the function into vertex form is another way to determine the vertex. You can do this when the leading coefficient is other than 1, but then locating the axis of symmetry from x = -b / 2a is probably easier.

y = x² - 10x + ___ + 24

Complete the square on the first three terms by taking 1/2 of -10 and squaring the result. Subtract 25 at the end to keep the equation in balance:

y = x² - 10x + 25 + 24 - 25

y = (x - 5)² - 1 (since the 1st 3 terms form a perfect square trinomial, write as a binomial squared)

In this form it is easy to identify the vertex as (5, -1).

Answer 3

Given a general quadratic equation of y=ax^2+bx+c, the x-coordinate of the vertex lies at -b/(2a).

Here, a=1, b=-10, and c=24.

So the x-coordinate of the vertex is -(-10)/2(1) = 10/2 = 5.

Moreover, the expression can be factored as (x-6)(x-4). Plugging in the x-coordinate of the vertex to get the y-value, we have (5-6)(5-4) = (-1)(1) = -1. So the vertex lies at (5,-1).

The general vertex form is y=a(x-h)^2+k, where (h,k) is the vertex and the "a" value is preserved from the above.

So the vertex form is f(x) = 1(x-5)^2-1 = (x-5)^2-1