McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
5. Quadratic Inequalities
Continue to next subchapter

Exercise 21 Page 211

Find the roots and use them to graph the related function.

{all real numbers}

Practice makes perfect
We will start by sketching the related quadratic function. To do so, we first need to identify the values of a, b, and c. y= 1x^2 + 8x + 16 We see that a= 1, b= 8, and c= 16. Let's substitute these values into the Quadratic Formula to find the roots of x^2+8x+16=0.
x=- b±sqrt(b^2-4ac)/2a
x=- 8±sqrt(8^2-4( 1)( 16))/2( 1)
Simplify right-hand side
x=- 8±sqrt(64-4(1)(16))/2(1)
x=- 8±sqrt(64-64)/2
x=- 8±sqrt(0)/2
x=- 8 ± 0/2
x=- 8/2
x=- 4

We see that there is only one root, x=- 4. The solution of the given quadratic inequality, x^2+8x+16≥0, consists of x-values for which the graph of the related quadratic function lies on and above the x-axis. The graph opens upward since a= 1 is greater than zero.

We see that the graph lies on and above the x-axis for all x-values. {all real numbers} ⇕ (- ∞, + ∞ )

Showing Our Work

Drawing the parabola using roots and vertex
We already know the root, which is - 4. To draw the graph, we will find the vertex, (- b2a,f(- b2a)). Let's start by calculating its x-coordinate. To do so, we will substitute a= 1 and b= 8 into - b2a.
- b/2a
- 8/2( 1)
- 8/2
- 4
Finally, to find the y-coordinate of the vertex, we will substitute - 4 for x in the related function.
y=x^2+8x+16
y=( - 4 )^2+8 ( - 4) +16
Simplify right-hand side
y=16+8(- 4) +16
y=16-32 +16
y=0
The vertex is (- 4,0). This point, along with the roots, is helpful to graph a parabola.