SOLUTION: If x can take any whole number value from 1 to 10, find the smallest possible value of x2-8x+25.

Question 1011338: If x can take any whole number value from 1 to 10, find the smallest possible value of x2-8x+25.
Found 2 solutions by mathmate, Theo:
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:
If x can take any whole number value from 1 to 10, find the smallest possible value of x2-8x+25.

Solution:
For
Three possible ways.

1. By trial and error (x=1 to 10).
We can form a table and find that at x=4, f(4)=16-32+25=9

2. By finding the vertex of the parabola, similar to completing the square.
f(x)=(x-4)^2+9, so at x=4, f(x)=9.

3. Using calculus.
f(x)=x^2-8x+25
f'(x)=2x-8=0 => x=4
f"(x)=2>0 => x=4 is a minimum
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the smallest possible value a quadratic in the form of ax^2 + bx + c = 0 would be at x = -b/2a.

your equation of x^2 -8x + 25 is already in that form when you set it equal to 0.

a = coefficient of x^2 term = 1
b = coefficient of x term = -8
c = coefficient of c term = 25

x = -b/2a = -(-8)/(2*1) = 8/2 = 4

when x = 4, y = x^2 - 8x + 25 becomes y = 4^2 - 8*4 + 25 which becomes 16 - 32 + 25 which becomes 9.

the smallest value of x^2-8x+25 would be equal to 9.

here's what the graph looks like.

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