Question

algebra 2 362403-6 fy - 1
benchmark test 2
what is the vertex of the graph of $f(x) = x^2 - 2x - 3$?
a. $(1, 2)$
b. $(1, 4)$
c. $(1, -4)$
d. $(2, 1)$

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 \( x = -\frac{b}{2a} \). In the function \( f(x)=x^{2}-2x - 3 \), \( a = 1 \), \( b=-2 \), \( c=-3 \).

Step2: Calculate x - coordinate of vertex

Substitute \( a = 1 \) and \( b=-2 \) into the formula \( x=-\frac{b}{2a} \). We get \( x =-\frac{-2}{2\times1}=\frac{2}{2} = 1 \).

Step3: Calculate y - coordinate of vertex

Substitute \( x = 1 \) into the function \( f(x)=x^{2}-2x - 3 \). So \( f(1)=(1)^{2}-2\times(1)-3=1 - 2 - 3=-4 \).

Answer:

C. \( (1, -4) \)