Question

find the y-intercept, the axis of symmetry, and the vertex of the graph of the function.
$f(x) = -3x^2 + 6x - 2$

the y-intercept is $(0, -2)$. (type an ordered pair.)
the axis of symmetry is $x = 1$. (simplify your answer. type an equation.)
the vertex is \boxed{}. (type an ordered pair.)

Explanation:

Step1: Recall vertex formula for parabola

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 from the axis of symmetry that \( x = 1 \) (since the axis of symmetry of a parabola \( y=ax^{2}+bx + c \) is \( x=-\frac{b}{2a} \)).

Step2: Find the y - coordinate of the vertex

To find the y - coordinate, we substitute \( x = 1 \) into the function \( f(x)=-3x^{2}+6x - 2 \).
Substitute \( x = 1 \) into \( f(x) \):
\( f(1)=-3(1)^{2}+6(1)-2 \)
First, calculate the powers and multiplications: \( -3\times1=-3 \), \( 6\times1 = 6 \)
Then, perform the addition and subtraction: \( f(1)=-3 + 6-2=1 \)

Answer:

\((1,1)\)