Y=x^2-6x+14
=x^2-6x+9-9+14
x^2-3x-3x+9 +5
x(x-3)-3(x-3)+5
(x-3)^2+5
here h=3
minimum value would be 5
Answer:
Quadrant III
Step-by-step explanation:
The attached picture shows graph of 4 such linear functions with the conditions given in the problem. ALL of them DO NOT pass through Quadrant III.
The graphs shown are of the functions:




<em>So, any linear function of the form
with
and
does not pass through Quadrant III. Answer choice 3 is correct.</em>
The last one is rational because it is a terminating decimal.
That menas that it we be negative because any thing that is negative is going bigger