Inequalities systems and graphs #2 - Q&A

Question 1 Graph the function rule. Explain why the graph is continuous or discrete. Graph: 𝑦 = − ∣ 𝑥 ∣ + 2 𝑦 = − ∣ 𝑥 ∣ + 2 y=−∣x∣+2y=−∣x∣+2 Answer 1 To graph the function rule 𝑦 = − ∣ 𝑥 ∣ + 2 y=−∣x∣+2, we choose several input values 𝑥 x and find the corresponding output values 𝑦 y. Then we create a table showing those inputs and outputs. Finally, we plot these pairs of input (x), output (y) on a rectangular coordinate system and join them. We find the output values 𝑦 y of the given function rule at 𝑥 = − 2 , − 1 , 0 , 1 , 2 x=−2,−1,0,1,2. Substituting: 𝑥 = − 2 x=−2 in the expression − ∣ 𝑥 ∣ + 2 −∣x∣+2, we get: 𝑦 = − ∣ − 2 ∣ + 2 = − 2 + 2 = 0. y=−∣−2∣+2=−2+2=0. Substituting 𝑥 = − 1 x=−1: 𝑦 = − ∣ − 1 ∣ + 2 = − 1 + 2 = 1. y=−∣−1∣+2=−1+2=1. Substituting 𝑥 = 0 x=0: 𝑦 = − ∣ 0 ∣ + 2 = 0 + 2 = 2. y=−∣0∣+2=0+2=2. Substituting 𝑥 = 1 x=1: 𝑦 = − ∣ 1 ∣ + 2 = − 1 + 2 = 1. y=−∣1∣+2=−1+2=1. Substituting 𝑥 = 2 x=2: 𝑦 = − ∣ 2 ∣ + 2 = − 2 + 2 = 0. y=−∣2∣+2=−2+2=0. The following table shows the values and the corresponding ordered pairs: 𝑥 : − 2 , − 1 , 0 , 1 , 2 𝑦 : 0 , 1 , 2 , 1 , 0 x:−2,−1,0,1,2y:0,1,2,1,0 Now plotting the ordered pairs on a rectangular coordinate system, we get the graph of the function rule 𝑦 = − ∣ 𝑥 ∣ + 2 y=−∣x∣+2. Graph attached. Question 2 Suggest a coordinate plane located at 𝑊 ( − 3 , 5 ) W(−3,5) and 𝑍 ( 1 , 1 ) Z(1,1) with 𝑊 ′ , 𝑍 ′ W ′ ,Z ′ . 𝑊 W becomes 𝑊 ′ ( 12 , 20 ) W ′ (12,20) and 𝑍 ′ ( 4 , 4 ) Z ′ (4,4), under a dilation centered at the origin. What is this dilation’s scale factor? Answer 1 If the ( 𝑥 , 𝑦 ) (x,y) point is dilated by a scale factor 𝑘 k centered at the origin, then the image has ( 𝑘 𝑥 , 𝑘 𝑦 ) (kx,ky) coordinates. Given: 𝑊 ( − 3 , 5 ) W(−3,5) 𝑊 ′ ( 12 , 20 ) W ′ (12,20) Since − 12 −12 and 20 20 are both products of − 3 −3 and 5 5 multiplied by 4 4, then the scale factor has to be 𝑘 = 4 k=4. To verify this, check the other points: 𝑍 ( 1 , 1 ) Z(1,1) 𝑍 ′ ( 4 , 4 ) . Z ′ (4,4). Since 4 4 is 1 1 multiplied by 4 4, it’s verified that the scale factor is 𝑘 = 4 k=4. Question 3 Graph equation: 2 𝑥 − 3 𝑦 = 6. 2x−3y=6. Answer 1 We can graph an equation by determining the 𝑥 x- and 𝑦 y-intercepts and then drawing the line through them. Set 𝑥 = 0 x=0 to find the 𝑦 y-intercept: 2 ( 0 ) − 3 𝑦 = 6 , 𝑦 = − 2. 2(0)−3y=6,y=−2. Set 𝑦 = 0 y=0 to find the 𝑥 x-intercept: 2 𝑥 − 3 ( 0 ) = 6 , 𝑥 = 3. 2x−3(0)=6,x=3. Graph and alternate explanations using slope-intercept form attached. Question 4 Graph the feasible region for the system of inequalities: 𝑦 > 4 𝑥 − 1 y>4x−1 𝑦 < 2 𝑥 + 3. y<2x+3. Answer 1 Step 1: Replace the inequality signs by equality signs and plot the two equations. 𝑦 = 4 𝑥 − 1 , 𝑦 = 2 𝑥 + 3. y=4x−1,y=2x+3. Check if ( 0 , 0 ) (0,0) satisfies the inequalities. For 𝑦 > 4 𝑥 − 1 y>4x−1: 0 > 4 ( 0 ) − 1 ⇒ 0 > − 1 (True for  ( 0 , 0 ) ) . 0>4(0)−1⇒0>−1(True for (0,0)). For 𝑦 < 2 𝑥 + 3 y<2x+3: 0 < 2 ( 0 ) + 3 ⇒ 0 < 3 (True for  ( 0 , 0 ) ) . 0<2(0)+3⇒0<3(True for (0,0)). Step 2: Plot the region of inequality and the region satisfying both the inequalities is the feasible region. The shaded region is the feasible region. Graph attached. Question 5 What are the two Special Inequalities? Answer 1 The two Special Inequalities are inequalities having no solution and inequalities having infinitely many solutions. Example: For 𝑥 + 𝑥 > 1 x+x>1: On simplification, we get: 𝑥 − 𝑥 + 𝑥 > 1 ⇒ 0 > 1 , x−x+x>1⇒0>1, which is not possible. Therefore, the inequality 𝑥 + 𝑥 > 1 x+x>1 has no solution. Consider 𝑥 x. On simplification, we get: 𝑥 − 𝑥 ⇒ 0 ≥ 1 , x−x⇒0≥1, which is true for every value of 𝑥 x. Therefore, the inequality 𝑥 x is special.