Exercise 109 Page 205

Boundary lines

To draw the boundary lines we will temporarily consider the inequalities related equation instead. We find a related equation by changing the inequality sign to an equals sign. |l|l| [-1.1em] Inequality & Equation [-0.5em] y≤ 4x+16 & y= 4x+16 [0.5em] [-0.9em] y>- 4/3x-4 & y=- 4/3x-4 [0.8em] Now we can graph the equations. Let the y-intercept be the first point and use the slope to find a second point. Notice that the first inequality is non-strict which means we keep the line solid. Let's show this with the first equation.

Let's also graph the second boundary line. Since this boundary line shows a strict inequality, it's not part of the solution set. Therefore, we will dash it.

Shading

To complete the inequalities we have to shade the correct side of the boundary line. We can do that by testing a point that is not on any of the lines. In this case, the easiest point we can choose is the origin. |c|c|c| [-1em] -3pt (x,y) -3pt & Inequality & Evaluate [0.2em] [-1em] -3pt ( 0, 0) -3pt & 0? ≤ 4( 0)+16 & 0≤ 16 ✓ [0.8em] [-0.8em] -3pt ( 0, 0) -3pt & 0? >- 4/3( 0)-4 & 0 > - 4 ✓ [0.8em] Since both inequalities are true for the origin, we have to shade the region of each boundary line that contains the origin.

Where the inequalities overlap, we have the solution set to the system of inequalities.

Boundary line/curve

Like in Part A, we will temporarily consider the inequalities related equation. |l|l| [-1.1em] Inequality & Equation [-0.5em] y< x^2-2x-3 & y= x^2-2x-3 [0.5em] [-0.9em] y≤3/4x+2 & y=3/4x+2 [0.8em] To graph the quadratic equation, we will substitute some x-values into the formula and calculate the corresponding y-value. |c|c|c| [-1em] x & x^2-2x-3 & y [0.2em] [-1em] -2 & ( -2)^2-2( -2)-3 & 5 [0.2em] [-1em] -1 & ( -1)^2-2( -1)-3 & 0 [0.2em] [-1em] 0 & 0^2-2( 0)-3 & -3 [0.2em] [-1em] 1 & 1^2-2( 1)-3 & -4 [0.2em] [-1em] 2 & 2^2-2( 2)-3 & -3 [0.2em] Now that we know a few points that fall on the curve, we can graph the boundary curve. Notice that the inequality is strict. Therefore, we will dash the curve.

Let's also graph the second equation. Like in Part A, this is a line which means we need at least two points to graph it. We can find these points by using the y-intercept and the slope. Since this boundary line shows a non-strict inequality we will keep it solid.

Shading

Like in Part A, we will test a point in each inequality that does not fall on any of the boundary lines. |c|c|c| [-1em] -3pt (x,y) -3pt & -2pt Inequality -2pt & Evaluate [0.2em] [-1em] -3pt ( 0, 0) -3pt & -2pt 0? < 0^2-2( 0)-3 -2pt & 0≮ -3 * [0.8em] [-0.8em] -3pt ( 0, 0) -3pt & -2pt 0? ≤3/4( 0)+2 -2pt & 0 ≤ 2 ✓ [0.8em] For the first inequality, we should shade the side of the boundary curve that does not contain the origin. For the second inequality, we should shade the side of the boundary line that contains the origin.

The region where both inequalities overlap shows the solution set to the system of inequalities.

Inequality:& y≥|x+2|-3 Equality:& y=|x+2|-3 This function contains an absolute value which means that y=|x| is its parent function. We can write an absolute value function in the following general format. y= a|x-h|+k [-1.1em] Horizontal Translation:& h Vertical Translation:& k Vertical Stretch:& a Notice that when we add to the input, we get a translation to the left. When we subtract from the output, we get a translation down. |c|l| -3pt Right-hand side -3pt & Transformation |x+2|-3& &Horizontal translation &to the left by 2 units. |x+2|-3& -3pt & Vertical translation down &by 3 units. -3pt Let's show these translations.

Like in previous parts, we will test a point in the original inequality to figure out which side of the boundary line we should shade.

Since the inequality is true for the origin, we should shade the side of the boundary line that contains this point.

Boundary line/curve

Like in previous parts, we will temporarily consider the related equation instead. |l|l| [-1.1em] Inequality & Equation [-0.8em] y≤ 1/2x+3 & y= 1/2x+3 [0.8em] [-0.6em] y≥ (x+1)^2-2 & y=(x+1)^2-2 [0.7em] Notice that both inequalities are non-strict. Therefore, both boundaries will be solid. Let's start by graphing the boundary line of the linear equation. It has a y-intercept of (0,3) and a slope of 12.

For the second equation, we notice that it resembles the graphing form of a parabola. By rewriting it so that it matches the graphing form exactly, we can identify the vertex and stretch factor.

y=(x+1)^2-2

▼

Write in graphing form

AddPar

Add parentheses

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

a=1a

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

Now we can identify the vertex and stretch factor of the quadratic function. y= 1(x-(-1))^2+(-2) [0.2em] [-1em] &Vertex: (-1,-2) &Stretch Factor: 1 Let's graph the vertex and the parabola's line of symmetry. This is a vertical line that runs through the vertex x-coordinate.

To graph the parabola, we need to identify more points that fall on the graph. By rewriting the right-hand side into standard form, we can identify the graph's y-intercept.

y=(x+1)^2-2

ExpandPosPerfectSquare

(a+b)^2=a^2+2ab+b^2

y=x^2+2x+1-2

SubTerm

Subtract term

y=x^2+2x-1

The function's constant is -1 which means it intercepts the y-axis at (0,-1). Let's add this to the coordinate plane.

We can find a third point by first measuring the horizontal distance between the y-intercept and the line of symmetry. Then, we will move the same distance to the left of the line of symmetry. Once we have three points, the parabola can be graphed accurately.

Shading

Like in previous parts, we will test a point in the inequalities that does not fall on any of the boundary lines. |c|c|c| [-1em] -3pt (x,y) -3pt & -2pt Inequality -2pt & Evaluate [0.2em] [-1em] -3pt ( 0, 0) -3pt & -2pt 0? ≤ 1/2( 0)+3 -2pt & 0≤ 3 ✓ [0.8em] [-0.8em] -3pt ( 0, 0) -3pt & -2pt 0? ≥( 0+1)^2-2 -2pt & 0 ≥ -1 ✓ [0.8em] For both inequalities, we should shade the region of the boundary line/curve that contains the origin.

The region that both inequalities share shows the solution set to the system of inequalities.