Question

graph the parabola and give its vertex, axis of symmetry, x-intercepts, and y-intercept.
$y = \frac{1}{5}x^2 - \frac{8}{5}x + \frac{11}{5}$

the vertex is \square. (type an ordered pair )

Explanation:

Step1: Recall vertex formula for parabola

For a quadratic function \( y = ax^2+bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). Here, \( a=\frac{1}{5} \), \( b =-\frac{8}{5} \), and \( c=\frac{11}{5} \).
Calculate the x - coordinate: \( x=-\frac{-\frac{8}{5}}{2\times\frac{1}{5}}=\frac{\frac{8}{5}}{\frac{2}{5}}=\frac{8}{5}\times\frac{5}{2} = 4 \)

Step2: Find the y - coordinate of the vertex

Substitute \( x = 4 \) into the equation \( y=\frac{1}{5}x^{2}-\frac{8}{5}x+\frac{11}{5} \)
\( y=\frac{1}{5}(4)^{2}-\frac{8}{5}(4)+\frac{11}{5}=\frac{16}{5}-\frac{32}{5}+\frac{11}{5}=\frac{16 - 32+11}{5}=\frac{-5}{5}=- 1 \)

Answer:

\((4, - 1)\)