Find the vertex form of the function. Then find each of the following.

​(A) Intercepts   
 (B) Vertex   
 ​(C) Maximum or minimum and what are the values  ​
(D) Range

s(x)= x^2-8x+7

1 answer

To find the vertex form of the function, we complete the square.

s(x) = x^2 - 8x + 7

First, we need to factor out the coefficient of the x^2 term, which is 1.

s(x) = (x^2 - 8x) + 7

Next, take half of the coefficient of the x term, which is -8 (-8/2 = -4), and square it (16).

s(x) = (x^2 - 8x + 16) + 7 - 16

Now, rewrite the expression inside parentheses as a binomial squared.

s(x) = (x - 4)^2 + 7 - 16

Simplify the expression within the parentheses.

s(x) = (x - 4)^2 - 9

Therefore, the vertex form of the function is s(x) = (x - 4)^2 - 9.

(A) To find the x-intercepts, we set s(x) = 0 and solve for x.

0 = (x - 4)^2 - 9

(x - 4)^2 = 9

Take the square root of both sides.

x - 4 = ±3

Solve for x.

x = 4 + 3 or x = 4 - 3

x = 7 or x = 1

The x-intercepts are (7, 0) and (1, 0).

(B) The vertex of the function is found by taking the opposite of the x-coordinate of the binomial squared and the y-coordinate is the constant term outside the parentheses.

Vertex = (4, -9)

(C) Since the coefficient of the x^2 term is positive (positive 1 in this case), the parabola opens upwards, which means the vertex is a minimum point. The minimum value of the function is the y-coordinate of the vertex.

Minimum value = -9

(D) The range of the function is all real numbers greater than or equal to the minimum value.

Range = y ≥ -9