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Precalculus Fifth Edition Mathematics for Calculus James Stewart Lothar Redlin Saleem Watson

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Precalculus Fifth Edition Mathematics for Calculus James Stewart Lothar Redlin Saleem Watson

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Coordinate Geometry

- The coordinate plane is the link between algebra and geometry.
- In the coordinate plane, we can draw graphs of algebraic equations.
- In turn, the graphs allow us to “see” the relationship between the variables in the equation.

The Coordinate Plane

- Points on a line can be identified with real numbers to form the coordinate line.
- Similarly, points in a plane can be identified with ordered pairs of numbers to form the coordinate plane or Cartesian plane.

Axes

- To do this, we draw two perpendicular real lines that intersect at 0 on each line.
- Usually,
- One line is horizontal with positive direction to the right and is called the x-axis.
- The other line is vertical with positive direction upward and is called the y-axis.

Origin & Quadrants

- The point of intersection of the x-axis and the y-axis is the origin O.
- Thetwo axes divide the plane into four quadrants, labeled I, II, III, and IV here.

Origin & Quadrants

- The points onthe coordinate axes are not assigned to any quadrant.

Ordered Pair

- Any point P in the coordinate plane can be located by a unique ordered pairof numbers (a, b).
- The first number a is called the x-coordinateof P.
- The second number b is called the y-coordinateof P.

Coordinates

- We can think of the coordinates of P as its “address.”
- They specify its location in the plane.

Coordinates

- Several points are labeled with their coordinates in this figure.

E.g. 1—Graphing Regions in the Coordinate Plane

- Describe and sketch the regions given by each set.
- {(x, y) | x≥ 0}
- {(x, y) | y = 1}
- {(x, y) | |y| < 1}

Example (a)E.g. 1—Regions in the Coord. Plane

- The points whose x-coordinates are 0 or positive lie on the y-axis or to the right of it.

Example (b)E.g. 1—Regions in the Coord. Plane

- The set of all points with y-coordinate 1 is a horizontal line one unit above the x-axis.

Example (c)E.g. 1—Regions in the Coord. Plane

- Recall from Section 1-7 that |y| < 1 if and only if –1 < y < 1
- So, the given region consists of those points in the plane whose y-coordinates lie between –1 and 1.
- Thus, the region consists of all points that lie between (but not on) the horizontal lines y = 1 and y = –1.

Example (c)E.g. 1—Regions in the Coord. Plane

- These lines are shown as broken lines here to indicate that the points on these lines do not lie in the set.

The Distance Formula

- We now find a formula for the distance d(A, B) between two points A(x1, y1) and B(x2, y2) in the plane.

The Distance Formula

- Recall from Section 1-1 that the distance between points a and b on a number line is: d(a, b) = |b – a|

The Distance Formula

- So, from the figure, we see that:
- The distance between the points A(x1, y1) and C(x2, y1) on a horizontal line must be |x2 – x1|.
- The distance between B(x2, y2) and C(x2, y1) on a vertical line must be |y2 – y1|.

The Distance Formula

- Triangle ABC is a right triangle.
- So, the Pythagorean Theorem gives:

Distance Formula

- The distance between the points A(x1, y1) and B(x2, y2) in the plane is:

E.g. 2—Applying the Distance Formula

- Which of the points P(1, –2) or Q(8, 9) is closer to the point A(5, 3)?
- By the Distance Formula, we have:

E.g. 2—Applying the Distance Formula

- This shows that d(P, A) < d(Q, A)
- So, P is closer to A.

The Midpoint Formula

- Now, let’s find the coordinates (x, y) of the midpoint M of the line segment that joins the point A(x1, y1) to the point B(x2, y2).

The Midpoint Formula

- In the figure, notice that triangles APM and MQB are congruent because:
- d(A, M) = d(M, B)
- The corresponding angles are equal.

The Midpoint Formula

- It follows that d(A, P) = d(M, Q).
- So, x – x1 = x2 – x

The Midpoint Formula

- Solving that equation for x, we get: 2x = x1 + x2
- Thus, x = (x1 + x2)/2
- Similarly, y = (y1 + y2)/2

Midpoint Formula

- The midpoint of the line segment from A(x1, y1) to B(x2, y2) is:

E.g. 3—Applying the Midpoint Formula

- Show that the quadrilateral with vertices P(1, 2), Q(4, 4), R(5, 9), and S(2, 7) is a parallelogram by proving that its two diagonals bisect each other.

E.g. 3—Applying the Midpoint Formula

- If the two diagonals have the same midpoint, they must bisect each other.
- The midpoint of the diagonal PR is:
- The midpoint of the diagonal QS is:

E.g. 3—Applying the Midpoint Formula

- Thus, each diagonal bisects the other.
- A theorem from elementary geometry states that the quadrilateral is therefore a parallelogram.

Equation in Two Variables

- An equation in two variables, such as y = x2 + 1, expresses a relationship between two quantities.

Graph of an Equation in Two Variables

- A point (x, y) satisfiesthe equation if it makes the equation true when the values for x and y are substituted into the equation.
- For example, the point (3, 10) satisfies the equation y = x2 + 1 because 10 = 32 + 1.
- However, the point (1, 3) does not, because 3 ≠ 12 + 1.

The Graph of an Equation

- The graphof an equation in x and y is:
- The set of all points (x, y) in the coordinate plane that satisfy the equation.

The Graph of an Equation

- The graph of an equation is a curve.
- So, to graph an equation, we:
- Plot as many points as we can.
- Connect them by a smooth curve.

E.g. 4—Sketching a Graph by Plotting Points

- Sketch the graph of the equation
- 2x – y = 3
- We first solve the given equation for y to get: y = 2x – 3

E.g. 4—Sketching a Graph by Plotting Points

- This helps us calculate the y-coordinates in this table.

E.g. 4—Sketching a Graph by Plotting Points

- Of course, there are infinitely many points on the graph—and it is impossible to plot all of them.
- Still, the more points we plot, the better we can imagine what the graph represented by the equation looks like.

E.g. 4—Sketching a Graph by Plotting Points

- We plot the points we found.
- As they appear to lie on a line, we complete the graph by joining the points by a line.

E.g. 4—Sketching a Graph by Plotting Points

- In Section 1-10, we verify that the graph of this equation is indeed a line.

E.g. 5—Sketching a Graph by Plotting Points

- Sketch the graph of the equation y = x2 – 2

E.g. 5—Sketching a Graph by Plotting Points

- We find some of the points that satisfy the equation in this table.

E.g. 5—Sketching a Graph by Plotting Points

- We plot these points and then connect them by a smooth curve.
- A curve with this shape is called a parabola.

E.g. 6—Graphing an Absolute Value Equation

- Sketch the graph of the equation y = |x|

E.g. 6—Graphing an Absolute Value Equation

- Again, we make a table of values.

E.g. 6—Graphing an Absolute Value Equation

- We plot these points and use them to sketch the graph of the equation.

x-intercepts

- The x-coordinates of the points where a graph intersects the x-axis are called the x-interceptsof the graph.
- They are obtained by setting y = 0 in the equation of the graph.

y-intercepts

- The y-coordinates of the points where a graph intersects the y-axis are called the y-interceptsof the graph.
- They are obtained by setting x = 0 in the equation of the graph.

E.g. 7—Finding Intercepts

- Find the x- and y-intercepts of the graph of the equation y = x2 – 2

E.g. 7—Finding Intercepts

- To find the x-intercepts, we set y = 0 and solve for x.
- Thus, 0 = x2 – 2 x2 = 2 (Add 2 to each side) (Take the sq. root)
- The x-intercepts are and .

E.g. 7—Finding Intercepts

- To find the y-intercepts, we set x = 0 and solve for y.
- Thus, y = 02 – 2 y = –2
- The y-intercept is –2.

E.g. 7—Finding Intercepts

- The graph of this equation was sketched in Example 5.
- It is repeated here with the x- and y-intercepts labeled.

Circles

- So far, we have discussed how to find the graph of an equation in x and y.
- The converse problem is to find an equation of a graph—an equation that represents a given curve in the xy-plane.

Circles

- Such an equation is satisfied by the coordinates of the points on the curve and by no other point.
- This is the other half of the fundamental principle of analytic geometry as formulated by Descartes and Fermat.

Circles

- The idea is that:
- If a geometric curve can be represented by an algebraic equation, then the rules of algebra can be used to analyze the curve.

Circles

- As an example of this type of problem, let’s find the equation of a circle with radius r and center (h, k).

Circles

- By definition, the circle is the set of all points P(x, y) whose distance from the center C(h, k) is r.
- Thus, P is on the circle if and only if d(P, C) = r

Circles

- From the distance formula, we have: (Square each side)
- This is the desired equation.

Equation of a Circle—Standard Form

- An equation of the circle with center (h, k) and radius r is: (x – h)2 + (y – k)2 = r2
- This is called the standard formfor the equation of the circle.

Equation of a Circle

- If the center of the circle is the origin (0, 0), then the equation is:x2 + y2 = r2

E.g. 8—Graphing a Circle

- Graph each equation.
- x2 + y2 = 25
- (x – 2)2 + (y + 1)2 = 25

Example (a)E.g. 8—Graphing a Circle

- Rewriting the equation as x2 + y2 = 52, we see that that this is an equation of:
- The circle of radius 5 centered at the origin.

Example (b)E.g. 8—Graphing a Circle

- Rewriting the equation as (x – 2)2 + (y + 1)2 = 52, wesee that this is an equation of:
- The circle of radius 5 centered at (2, –1).

E.g. 9—Finding an Equation of a Circle

- Find an equation of the circle with radius 3 and center (2, –5).
- (b) Find an equation of the circle that has the points P(1, 8) and Q(5, –6) as the endpoints of a diameter.

Example (a)E.g. 9—Equation of a Circle

- Using the equation of a circle with r = 3, h = 2, and k = –5, we obtain: (x – 2)2 + (y + 5)2 = 9

Example (b)E.g. 9—Equation of a Circle

- We first observe that the center is the midpoint of the diameter PQ.
- So,by the Midpoint Formula, the center is:

Example (b)E.g. 9—Equation of a Circle

- The radius r is the distance from P to the center.
- So, by the Distance Formula,r2 = (3 – 1)2 + (1 – 8)2 = 22 + (–7)2 = 53

Example (b)E.g. 9—Equation of a Circle

- Hence, the equation of the circle is: (x – 3)2 + (y – 1)2 = 53

Equation of a Circle

- Let’s expand the equation of the circle in the preceding example.
- (x – 3)2 + (y – 1)2 = 53 (Standard form)
- x2 – 6x + 9 + y2 – 2y + 1 = 53 (Expand the squares)
- x2 – 6x +y2 – 2y = 43 (Subtract 10 to get the expanded form)

Equation of a Circle

- Suppose we are given the equation of a circle in expanded form.
- Then, to find its center and radius, we must put the equation back in standard form.

Equation of a Circle

- That means we must reverse the steps in the preceding calculation.
- To do that, we need to know what to add to an expression like x2 – 6x to make it a perfect square.
- That is, we need to complete the square—as in the next example.

E.g. 10—Identifying an Equation of a Circle

- Show that the equation x2 + y2 + 2x – 6y + 7 = 0 represents a circle.
- Find the center and radius of the circle.

E.g. 10—Identifying an Equation of a Circle

- First, we group the x-terms and y-terms.
- Then, we complete the square within each grouping.
- We complete the square for x2 + 2x by adding (½ ∙ 2)2 = 1.
- We complete the square for y2 – 6y by adding [½ ∙ (–6)]2 = 9.

E.g. 10—Identifying an Equation of a Circle

- Comparing this equation with the standard equation of a circle, we see that: h = –1, k = 3, r =
- So, the given equation represents a circle with center (–1, 3) and radius .

Symmetry

- The figure shows the graph of y = x2
- Notice that the part of the graph to the left of the y-axis is the mirror image of the part to the right of the y-axis.

Symmetry

- The reason is that, if the point (x, y) is on the graph, then so is (–x, y), and these points are reflections of each other about the y-axis.

Symmetric with Respect to y-axis

- In this situation, we say the graph is symmetric with respect to the y-axis.

Symmetric with Respect to x-axis

- Similarly, we say a graph is symmetric with respect to the x-axis if, whenever the point (x, y) is on the graph, then so is (x, –y).

Symmetric with Respect to Origin

- A graph is symmetric with respect to the originif, whenever (x, y) is on the graph, so is (–x, –y).

Using Symmetry to Sketch a Graph

- The remaining examples in this section show how symmetry helps us sketch the graphs of equations.

E.g. 11—Using Symmetry to Sketch a Graph

- Test the equation x = y2for symmetry and sketch the graph.

E.g. 11—Using Symmetry to Sketch a Graph

- If y is replaced by –y in the equation x = y2, we get: x = (–y)2(Replace y by –y)x = y2(Simplify)
- So, the equation is unchanged.
- Thus, the graph is symmetric about the x-axis.

E.g. 11—Using Symmetry to Sketch a Graph

- However, changing x to –x gives the equation –x = y2
- This is not the same as the original equation.
- So, the graph is not symmetric about the y-axis.

E.g. 11—Using Symmetry to Sketch a Graph

- We use the symmetry about the x-axis to sketch the graph.
- First, we plot points just for y > 0.

E.g. 11—Using Symmetry to Sketch a Graph

- Then, we reflect the graph in the x-axis.

E.g. 12—Using Symmetry to Sketch a Graph

- Test the equation y = x3 – 9xfor symmetry and sketch its graph.

E.g. 12—Using Symmetry to Sketch a Graph

- If we replace x by –x and y by –y, we get: –y = (–x3) – 9(–x) –y = –x3 + 9x(Simplify)y = x3 – 9x(Multiply by –1)
- So, the equation is unchanged.
- This means that the graph is symmetric with respect to the origin.

E.g. 12—Using Symmetry to Sketch a Graph

- First, we plot points for x > 0.

E.g. 12—Using Symmetry to Sketch a Graph

- Then, we use symmetry about the origin.

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