Find the minimum value of $x^2-6x+10$ for real $x$.
A0
B1
C2
D4
Answer & Solution
Correct answer: B. 1
Complete the square: $x^2-6x+10=(x-3)^2+1$. Since $(x-3)^2\ge 0$ for all real $x$, the minimum value occurs when $(x-3)^2=0$, that is at $x=3$. Therefore the minimum value is $1$.
Related questions
Which of the following is a linear equation in one variable?A number divided by 2 is 5 less than the number itself. What is the number?The sum of three consecutive integers is 24. What are the integers?For the pair of linear equations $3x+2y=12$ and $-x+y=3$, what does the graphical solutionUsing the graph of $3x+2y=12$ and $-x+y=3$, what is the solution of the system?
![](httpsSolve the system by elimination: $x+y=7$ and $3x+2y=10$.A system of two linear equations in two variables has infinitely many solutions when whichWhat is the discriminant of the quadratic equation $Ax^2+Bx+C=0$?