McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
1. Graphing Quadratic Functions
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Exercise 50 Page 100

Practice makes perfect
a

For a quadratic function y=ax^2+bx+c, the y-coordinate of the vertex is the minimum value of the function when a>0.

Let's identify the values of a, b, and c in the given quadratic function. y=2x^2+4x+6 ⇕ y= 2x^2+ 4x+ 6 We can see above that a= 2, b= 4, and c= 6. We will now use these values to find the desired information. Since a= 2 is greater than 0, the parabola will open upwards. This means it will have a minimum value.

b

The minimum value is given by substituting - b2a for x. Before we find the value of the function at this point, we need to substitute a= 2 and b= 4 in - b2a.

- b/2a
â–¼
Substitute values and evaluate
- 4/2( 2)
- 4/4
- 1

Now we have to substitute - 1 for x in the given equation.

y = 2x^2+4x+6
y = 2( - 1)^2+4( - 1)+6
â–¼
Simplify right-hand side
y = 2(1)+4(- 1)+6
y = 2(1)-4+6
y = 2-4+6
y = 4

This tells us that the minimum value of the function is 4.

c

Unless there are any specified restrictions on the x-values, the domain of a quadratic function is all real numbers. Therefore, the domain of this function is all real numbers. Furthermore, since a= 2 is greater than 0, the range is all values greater than or equal to the minimum value, 4.

Domain:& {All real numbers} Range:& {y | y ≥ 4}