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# Precalculus Fifth Edition Mathematics for Calculus James Stewart  Lothar Redlin  Saleem Watson - PowerPoint PPT Presentation

Precalculus Fifth Edition Mathematics for Calculus James Stewart  Lothar Redlin  Saleem Watson. Fundamentals. 1. Coordinate Geometry. 1.8. Coordinate Geometry. The coordinate plane is the link between algebra and geometry.

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Fifth Edition

Mathematics for Calculus

James StewartLothar RedlinSaleem Watson

• 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.

• 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.

• 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.

• 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.

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

• 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.

• We can think of the coordinates of P as its “address.”

• They specify its location in the plane.

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

• Describe and sketch the regions given by each set.

• {(x, y) | x≥ 0}

• {(x, y) | y = 1}

• {(x, y) | |y| < 1}

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.

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.

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.

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.

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

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

• 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|.

• Triangle ABC is a right triangle.

• So, the Pythagorean Theorem gives:

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

• 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:

• This shows that d(P, A) < d(Q, A)

• So, P is closer to A.

• 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).

• In the figure, notice that triangles APM and MQB are congruent because:

• d(A, M) = d(M, B)

• The corresponding angles are equal.

• It follows that d(A, P) = d(M, Q).

• So, x – x1 = x2 – x

• Solving that equation for x, we get: 2x = x1 + x2

• Thus, x = (x1 + x2)/2

• Similarly, y = (y1 + y2)/2

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

• 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.

• 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:

• Thus, each diagonal bisects the other.

• A theorem from elementary geometry states that the quadrilateral is therefore a parallelogram.

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

• 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 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 is a curve.

• So, to graph an equation, we:

• Plot as many points as we can.

• Connect them by a smooth curve.

• Sketch the graph of the equation

• 2x – y = 3

• We first solve the given equation for y to get: y = 2x – 3

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

• 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.

• 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.

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

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

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

• We plot these points and then connect them by a smooth curve.

• A curve with this shape is called a parabola.

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

• Again, we make a table of values.

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

• 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.

• 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.

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

• 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 .

• To find the y-intercepts, we set x = 0 and solve for y.

• Thus, y = 02 – 2 y = –2

• The y-intercept is –2.

• The graph of this equation was sketched in Example 5.

• It is repeated here with the x- and y-intercepts labeled.

• 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.

• 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.

• 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.

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

• 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

• From the distance formula, we have: (Square each side)

• This is the desired equation.

• 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.

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

• Graph each equation.

• x2 + y2 = 25

• (x – 2)2 + (y + 1)2 = 25

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.

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).

• 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.

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

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:

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

E.g. 9—Equation of a Circle

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

• 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)

• 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.

• 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.

• Show that the equation x2 + y2 + 2x – 6y + 7 = 0 represents a circle.

• Find the center and radius of the 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.

• 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 .

• 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.

• 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.

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

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

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

• Test the equation x = y2for symmetry and sketch the 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.

• 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.

• We use the symmetry about the x-axis to sketch the graph.

• First, we plot points just for y > 0.

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

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

• First, we plot points for x > 0.

• Then, we use symmetry about the origin.