SOLUTION: a)Express x^2 -10x + 32 in the form (x+a)^2 + b ,giving the values of a and b. b)Sketch the graph of y= x^2 -10x + 32, marking the coordinates of the minimum point on the curve.

Question 1084368: a)Express x^2 -10x + 32 in the form (x+a)^2 + b ,giving the values of a and b.
b)Sketch the graph of y= x^2 -10x + 32, marking the coordinates of the minimum point on the curve.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
start with x^2 - 10x + 32

put parentheses around (x^2 - 10x) to get (x^2 - 10x) + 32

take 1/2 the coefficient of the x term to get ((x-5)^2 - 25) + 32

what this says is that (x-5)^2 - 25 is equal to x^2 - 10x.

to see if that's true, multiply (x-5) * (x-5).

you will get x^2 - 10x + 25

since you want to wind up with x^2 - 10x, you have to subtract 25.

that's why (x-5)^2 - 25 is equal to x^2 -10x.

therefore in the expression of (x^2 + 10x) + 32, you replace (x^2 + 10x) with ((x-5)^2 - 25) to get ((x-5)^2 - 25) + 32.

when you remove parentheses, this becomes (x-5)^2 - 25 + 32 which becomes (x-5^2 + 7 after you combine like terms.

the graph of (x-5)^2 + 7 and the graph of x^2 - 10x + 32 are shown below:

$$$

you will see that both equations generate the same exact graph.

this means those two equations are identical.

the graph also shows you the minimum point on the graph is (5,7).

this means that the value of y doesn't go below 7, and the value of y is 7 when x = 5.

from (x-5)^2 + 7, you can see that the graph can go above y = 7 but can't go below y = 7 because (x-5)^2 is at a minimum value when x = 5.

from x^2 - 10x + 32, you would find the minimum point at x = -b/2a

that becomes x = 10/2 = 5

when x = 5, x^2 - 10x + 32 becomes 25 - 50 + 32 = -25 + 32 = 7

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