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
