Question

charla wants to determine the vertex of the function f(x) = x2 – 18x 60 by changing the function into vertex form. which statement about the vertex of the function is true? the x-coordinate of the vertex is greater than the y-coordinate. the x-coordinate of the vertex is negative. the y-coordinate of the vertex is greater than the y-intercept. the y-coordinate of the vertex is positive.

Answer 1

f(x) = x^2 - 18x + 60
f(x) = x^2 - 18x + 81 + 60 - 81
f(x) = (x - 9)^2 - 21
Vertex = (9, -21)
The x-coordinate of the vertex is greater than the y-coordinate.

Answer 2

Answer:

the x-coordinate of the vertex is greater than the y-coordinate

Step-by-step explanation:

we have

$f(x) = x^2 - 18x + 60$

we know that

The equation of a vertical parabola into vertex form is equal to

$f(x)=a(x-h)^{2} +k$

where

(h,k) is the vertex of the parabola

Convert into vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-60= x^2- 18x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)-60+81= x^2- 18x+81$

$f(x)+21= x^2- 18x+81$

Rewrite as perfect squares

$f(x)+21= (x-9)^2$

$f(x)= (x-9)^2-21$

The vertex is the point $(9,-21)$

Statements

case A) the x-coordinate of the vertex is greater than the y-coordinate

The statement is true -------> $9> -21$

case B) the x-coordinate of the vertex is negative

The statement is false ------> the x-coordinate of the vertex is positive

case C) the y-coordinate of the vertex is greater than the y-intercept

The statement is false-------> The vertex is a minimum (parabola open upward)

case D) the y-coordinate of the vertex is positive

The statement is false------->the y-coordinate of the vertex is negative