Practicing Success
Practicing Success
CUET Preparation Today
If x is real, the minimum value of $x^2-8x+17$ is : |
-1 0 1 2 |
1 |
The correct answer is Option (3) → 1 $y=x^2-8x+17$ $y=(x-4)^2+1$ min $(x-4)^2=0$ $y_{min}=1$ |
