Exercise 21 Page 286

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.

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

SubstituteII

a= 1, b= 8

- 8/2( 1)

Multiply

Multiply

- 8/2

CalcQuot

Calculate quotient

- 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

Substitute

x= - 4

y=( - 4 )^2+8 ( - 4) +16

▼

Simplify right-hand side

CalcPow

Calculate power

y=16+8(- 4) +16

MultPosNeg

a(- b)=- a * b

y=16-32 +16

AddSubTerms

Add and subtract terms

y=0

The vertex is (- 4,0). This point, along with the roots, is helpful to graph a parabola.