Exercise 5 Page 210

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 + 15 We see that a= 1, b= 8, and c= 15. Let's substitute these values into the Quadratic Formula to find the roots of x^2+8x+15=0.

Now we can calculate the first root using the positive sign and the second root using the negative sign.

The solution of the given quadratic inequality, x^2+8x+15 < 0, consists of x-values for which the graph of the related quadratic function lies below the x-axis. The graph opens upwards since a= 1 is greater than zero.

We see that the graph lies below the x-axis between x=- 5 and x=- 3. { x| - 5 < x < - 3 } ⇕ (- 5, - 3 )

Showing Our Work

Drawing the parabola using roots and vertex

We already know the roots, which are - 5 and - 3. 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)

IdPropMult

Identity Property of Multiplication

- 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+15.

y=x^2+8x+15

Substitute

x= - 4

y=(- 4)^2+8( - 4) +15

▼

Simplify right-hand side

NegBaseToPosBase

(- a)^2=a^2

y=(4)^2+8(- 4)+15

CalcPow

Calculate power

y=16+8(- 4)+15

MultPosNeg

a(- b)=- a * b

y=16-32+15

AddSubTerms

Add and subtract terms

y=- 1

The vertex is (- 4 ,- 1). This point, along with the roots, is helpful to graph a parabola.