McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
Study Guide and Review
Continue to next subchapter

Exercise 11 Page 158

Practice makes perfect
a Let's identify the values of a, b, and c in the given quadratic function.

y=x^2-4x+4 ⇕ y= 1x^2+( -4)x+ 4 We can see above that a= 1, b= -4, and c= 4. Since a>0, the parabola will have a minimum value.

b Since a= 1 is greater than 0, the parabola will open upwards. This means it will have a minimum value. To find it, we will evaluate the given function at x=- b2 a. Before we find the value of the function at this point, we need to substitute a= 1 and b= -4 in - b2 a.

- b/2a
â–Ľ
Substitute values and evaluate
- -4/2( 1)
- -4/2
4/2
2
Now we have to calculate the value of the function at x=2. To do so, we will substitute 2 for x in the given equation.
y=x^2-4x+4
y= 2^2-4( 2)+4
â–Ľ
Simplify right-hand side
y=4-4(2)+4
y=4-8+4
y=0
This tells us that the minimum value of the function is 0.
c Unless there are any specified restrictions on the x-values, the domain of a quadratic function is all real numbers. Furthermore, since the minimum value of the function is y=0, the range is all values greater than or equal to 0.

Domain:& All real numbers Range:& {y|y ≥ 0}