Exercise 7 Page 210

We will start by rearranging the inequality. 0 ≥ 2x^2-4x+1 ⇔ 2x^2-4x+1 ≤ 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=2x^2-4x+1 ⇔ y= 2x^2+( - 4)x+ 1 We see that a= 2, b= - 4, and c= 1. Let's substitute these values into the Quadratic Formula to find the roots of 2x^2-4x+1=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, 2x^2-4x+1≤0, consists of x-values for which the graph of the related quadratic function lies on and below the x-axis. The graph opens upwards since a= 2 is greater than zero.

We see that the graph lies on and below the x-axis at about x ≥ 0.29 and x ≤ 1.71. { x| 0.29 ≤ x ≤ 1.71 } ⇕ [0.29,1.71]

Showing Our Work

Drawing the parabola using roots and vertex

We already know the roots, which are around 0.29 and 1.71. 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= 2 and b= - 4 into - b2a.

- b/2a

SubstituteII

a= 2, b= - 4

- - 4/2( 2)

Multiply

Multiply

- - 4/4

RemoveNegFracAndNum

- - a/b= a/b

4/4

CalcQuot

Calculate quotient

1

Finally, to find the y-coordinate of the vertex, we will substitute 1 for x in the related function y=2x^2-4x+1.

y=2x^2-4x+1

Substitute

x= 1

y=2(1)^2-4 ( 1) +1

▼

Simplify right-hand side

BaseOne

1^a=1

y=2(1)-4(1)+1

Multiply

Multiply

y=2-4+1

AddSubTerms

Add and subtract terms

y=- 1

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