McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Quadratic Inequalities
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Exercise 5 Page 210

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

{ x| - 5 < x < - 3 }

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 + 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.
x=- b±sqrt(b^2-4ac)/2a
x=- 8±sqrt(( 8)^2-4( 1)( 15))/2( 1)
Simplify right-hand side
x=- 8±sqrt(64-4(1)(15))/2(1)
x=- 8 ± sqrt(64-60)/2
x=- 8 ± sqrt(4)/2
x=- 8 ± 2/2
Now we can calculate the first root using the positive sign and the second root using the negative sign.
x=- 8 ± 2/2
x=- 8 + 2/2 x=- 8 - 2/2
x=- 8/2+2/2 x=- 10/2-2/2
x= - 3 x=- 5

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
- 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+15.
y=x^2+8x+15
y=(- 4)^2+8( - 4) +15
Simplify right-hand side
y=(4)^2+8(- 4)+15
y=16+8(- 4)+15
y=16-32+15
y=- 1
The vertex is (- 4 ,- 1). This point, along with the roots, is helpful to graph a parabola.