Question

find the y-intercept, the axis of symmetry, and the vertex of the graph of the function.
$f(x)=0.6x^{2}-3.6x$

the y-intercept is (0,0). (type an ordered pair.)
the axis of symmetry is $x = 3$. (simplify your answer. type an equation.)
the vertex is \boxed{}. (type an ordered pair.)

Explanation:

Step1: Recall the vertex formula

For a quadratic function \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \). We already know the axis of symmetry is \( x = 3 \), so we can substitute \( x = 3 \) into the function \( f(x)=0.6x^2 - 3.6x \) to find the y-coordinate of the vertex.

Step2: Substitute x = 3 into the function

Substitute \( x = 3 \) into \( f(x) = 0.6x^2 - 3.6x \):
\( f(3)=0.6\times(3)^2 - 3.6\times(3) \)
First, calculate \( (3)^2 = 9 \), then \( 0.6\times9 = 5.4 \) and \( 3.6\times3 = 10.8 \).
So, \( f(3)=5.4 - 10.8=- 5.4 \)

Answer:

\((3, - 5.4)\)