McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Quadratic Inequalities
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Exercise 4 Page 210

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

{ x| x < 1 or x > 4 }

Practice makes perfect
We will start by rearranging the inequality. 0 < x^2-5x+4 ⇔ x^2-5x+4 > 0 Now, we are ready to sketch the related quadratic function. To do so, we first need to identify the values of a, b, and c. y=x^2-5x+4 ⇔ y= 1x^2+( - 5)x+ 4 We see that a= 1, b= - 5, and c= 4. Let's substitute these values into the Quadratic Formula to find the roots of x^2-5x+4=0.
x=- b±sqrt(b^2-4ac)/2a
x=- ( - 5)±sqrt(( - 5)^2-4( 1)( 4))/2( 1)
Simplify right-hand side
x=5±sqrt((- 5)^2-4(1)(4))/2(1)
x=5±sqrt(25-4(1)(4))/2(1)
x=5 ± sqrt(25-16)/2
x=5 ± sqrt(9)/2
x=5 ± 3/2
Now we can calculate the first root using the positive sign and the second root using the negative sign.
x=5 ± 3/2
x=5 + 3/2 x=5 - 3/2
x=8/2 x=2/2
x=4 x=1

The solution of the given quadratic inequality, x^2-5x+4>0, consists of x-values for which the graph of the related quadratic function lies above the x-axis.The graph opens upwards since a= 1 is greater than zero.

We see that the graph lies above the x-axis at x<1 and x >4. { x| x < 1 or x > 4 } ⇕ (- ∞, 1) ⋃ (4, ∞ )

Showing Our Work

Drawing the parabola using roots and vertex
We already know the roots, which are 1 and 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= - 5 into - b2a.
- b/2a
- - 5/2( 1)
- - 5/2
5/2
Finally, to find the y-coordinate of the vertex, we will substitute 52 for x in the related function y=x^2-5x+4.
y=x^2-5x+4
y=( 5/2 )^2 - 5 ( 5/2 ) +4
Simplify right-hand side
y=( 25/4 )-5( 5/2 ) +4
y=25/4 -25/2+4
y=25/4 -50/4+4
y=25/4 -50/4+16/4
y=- 9/4
y=2 14
The vertex is ( 52,2 14 ). This point, along with the roots, is helpful to graph a parabola.