Question

find the coordinates of the vertex of the following parabola algebraically. write your answer as an (x,y) point. y = x² - 12x + 32

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 \( x=-\frac{b}{2a} \). In the equation \( y = x^{2}-12x + 32 \), we have \( a = 1 \), \( b=- 12 \) and \( c = 32 \).

Step2: Calculate x - coordinate of vertex

Substitute \( a = 1 \) and \( b=-12 \) into the formula \( x =-\frac{b}{2a} \).
\( x=-\frac{-12}{2\times1}=\frac{12}{2} = 6 \)

Step3: Calculate y - coordinate of vertex

Substitute \( x = 6 \) into the equation \( y=x^{2}-12x + 32 \).
\( y=(6)^{2}-12\times6 + 32=36-72 + 32=36 + 32-72=68 - 72=-4 \)

Answer:

\((6,-4)\)