Answer:
there's a minimum at (4, -1)
Step-by-step explanation:
Please use " ^ " to denote exponentiation: y= x^2 - 8x+ 15.
To "complete the square," take half of the coefficient of x (which is -8). Square this result, obtaining 16.
In y= x^2 - 8x+ 15, add 16, and then subtract 16, between -8x and +15:
y = x^2 - 8x + 16 - 16 + 15
This becomes:
y = (x - 4)^2 -1
Reading off the coordinates of the vertex, we get (4, -1). Because the coefficient of the (x - 4)^2 term is positive, we know .there's a minimum at (4, -1)
