Question

estimate the coordinates of the vertex of the graph of $f(x) = 1.25x^2 - 2x - 1$ below. then explain how to find the exact coordinate. the x - coordinate of the vertex is approximately \boxed{ }, and the y - coordinate of the vertex is approximately \boxed{ }. (round to the nearest integer as needed.)

Explanation:

Step1: Find x - coordinate of vertex

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 \( f(x) = 1.25x^{2}-2x - 1 \), \( a = 1.25 \) and \( b=- 2 \).
So, \( x=-\frac{-2}{2\times1.25}=\frac{2}{2.5} = 0.8\approx1 \) (rounded to the nearest integer).

Step2: Find y - coordinate of vertex

Substitute \( x = 0.8 \) into the function \( f(x)=1.25x^{2}-2x - 1 \).
\( f(0.8)=1.25\times(0.8)^{2}-2\times0.8 - 1 \)
First, calculate \( (0.8)^{2}=0.64 \), then \( 1.25\times0.64 = 0.8 \)
\( 2\times0.8 = 1.6 \)
So, \( f(0.8)=0.8-1.6 - 1=-1.8\approx - 2 \) (rounded to the nearest integer).

Answer:

The x - coordinate of the vertex is approximately \( 1 \), and the y - coordinate of the vertex is approximately \( - 2 \)