Survey of Algebra and Probability: MAT 101

The point \(A\) is in \(Q1\) and point \(B\) is in \(Q2\text{.}\) Point \(C\) is not in a quadrant but is on the \(y\)-axis.

Tracing from the origin, point \(A\) is right \(1\text{,}\) up \(4\text{.}\) This becomes \(A(1, 4)\text{.}\) Point \(B\) is left \(5\text{,}\) up \(3\text{.}\) Left is backwards or negative so we have \(B( - 5, 3)\text{.}\) \(C\) is straight down 2 units. There is no left or right. This means we go right zero so the point is \(C(0,-2)\text{.}\)

The first point, \(A\) is at \((3, 2)\text{.}\) This means \(x=3\)(right 3) and \(y =2\)(up 2). Following these instructions, starting from the origin, we get our point. Note that \(A\) is in Q1.

The second point, \(B( - 2, 1)\text{,}\) is left 2 (negative moves backwards), up 1. This is also illustrated on the graph. Note that \(B\) is in Q2.

The third point, \(C(3,-4)\) is right 3, down 4 (negative moves backwards). Note that \(C\) is in Q4.

The fourth point, \(D(-2,-3)\) is left 2, down 3 (both negative, both move backwards). Note that \(D\) is in Q3.

Next is \(E(-3, 0)\text{.}\) This is left 3 (negative is backwards), and up zero, right on the \(x\)-axis. Note that \(E\) is on the \(x\)-axis.

Then is \(F(0, 2)\text{.}\) This is right zero, and up two, directly on the \(y\)-axis. Note that \(F\) on the \(y\)-axis.

Finally is \(G(0, 0)\text{.}\) This point has no movement. Thus the point is the origin.

Our Solution \(\checkmark\)

We will plug each given \(P\) value in the equation and solve for \(Q\)

Find each \(Q\)-coordinate by replacing \(P\) with the given value in the equation: \(Q = P^2-3\)

\(P=-3\mapsto Q =(-3)^2-3=9-3=6\) \(P=-1\mapsto Q =(-1)^2-3=1-3=-2\) \(P=0\mapsto Q =(0)^2-3=0-3=-3\) \(P=1\mapsto Q =(1)^2-3=1-3=-2\) \(P=5\mapsto Q =(5)^2-3=25-3=22\checkmark\)

We will test three values for \(x\text{.}\) We chose three integers easy to work, but any three values can be used.

We find the \(y\)-value for each \(x\)-value:

\(x=-1\mapsto y =2(-1)-3=-2-3=-5\) \(x=0\mapsto y =2(0)-3=0-3=-3\) \(x=1\mapsto y =2(1)-3=2-3=-1\)

We plot each point.

In Section 3.B, we will see thatall solutionjs of this particular equation lie on a line. Thus, once the points are plotted on the graph, we connect the dots to make a line.

The graph is Our Solution.\(\checkmark\)

We will test three values for \(x\text{.}\) WE chose three integers easy to work with, but any three values can be used.

Our completed table.

Graph points and connect dots.

Our Solution \(\checkmark\)

We will test five values for \(x\text{.}\) Any five can be used

Evaluate each by replacing \(x\) with the given value:

\(x=-2\mapsto y =|-2|=2\)

\(x=-1\mapsto y =|-1|=1\)

\(x=0\mapsto y =\lvert 0\rvert =0\)

\(x=1\mapsto y =\lvert 1\rvert =1\)

\(x=2\mapsto y =\lvert 2\rvert =2\)

We plot each point.

The graph is our solution.

Notice the graph of \(y=\lvert x\rvert\) lies entirely above the \(y\)-axis except at the origin. \(\checkmark\)

The \(y\)-intercept \((0,2)\) is labeled on the graph of the equation.

The \(x\)-intercept \(\left(-\frac{2}{3} , 0\right)\) is labeled on the graph of the equation.

The \(Q\)-intercept \(\left(0, \frac{8}{5}\right)\) is labeled on the graph of the equation.

The common point is \((1,3).\)

We say the graphs intersect at common points.

The graphs intersect at \(t=1.\checkmark\)

Note: The location of the common point is often given by the value from the horizontal axis and not by the entire point.

Using the graph, we see the red \(y=\lvert x+1\rvert\) and blue line \(y=3\) intersect at the two points \((-4,3)\) and \((2,3)\text{.}\)

Thus, \(\lvert x+1\rvert =3\) at \(x=-4\) or \(x=2.\checkmark\)

Using the graph, we see the red \(y=\lvert x+1\rvert\) lies below the blue line \(y=3\) for \(x\) values between \(-4\) and \(2\text{.}\)

Thus, \(\lvert x+1\rvert \lt 3\) for all \(x\) in the interval \((-4,2).\checkmark\)

Using the graph, we see the red \(y=\lvert x+1\rvert\) lies above the blue line \(y=3\) for \(x\) values less than \(-4\) or above \(2\text{.}\)

Thus, \(\lvert x+1\rvert \gt 3\) for all \(x\) in the compound interval \((-\infty,-4)\cup (2,\infty).\checkmark\)