Merced College; Don Power
INTERMEDIATE ALGEBRA - CH 10, LECTURE
McKeague
9.1 The Circle
Distance formula: Apply Pythagorean Thm to (x1,y1) and (x2 and y2). Result...
Ex: Find distance between (-3,-5) and (-7,2)
-- by the distance formula. Result: sqrt (42+72) = sqrt(65)
-- by plotting the points, sketching a triangle, and using the Pythagorean Theorem.
Intuitive check: how far apart are the x's? the y's?
Note: distance is never negative.
Circle:
Ex: Find circle with center (-2,5) and radius 6:
Use dist. formula, with one point fixed (center), the other variable (x,y), and distance = radius.
Note: to solve for y, square both sides; square root principle gives ± (upper, lower halves)
Most convenient (standard) form is (x-h)2 + (y-k)2 = r2 (Text uses (a,b) for center)
Result: (x+2)2 + (y-5)2 = 36
Simplified form for center at origin: h and k are 0.
Ex: Write equation for circle with radius sqrt(5), radius at origin. Result: x2 + y2 = 5
Compare the two
forms: a relation in which x and y are
replaced by (x-h) and (y-k)
has the same graph (size and
shape) as the simpler relation, but the origin is shifted to (h,k)
Finding center and radius if the equation is given:
From standard form.
From form ax2 + ab2 + cx + dy = constant: complete the square on both x and y.
Ex: x2 + y2 +12x -6y = 4 (Note: the center will not be 2)
x2 + 12x + 36 +y2 - 6y + 9 = 4 + 36 + 9 = 49. Result: center (6,-3), r = 7
9.2 Ellipses and Hyperbolas
Ellipse centered at origin: x2 / a2 + y2 / b2 = 1
x-intercepts: ±a Why? Set y=0, clear fraction.
y-intercepts: ±b Why? Set x=0, clear fraction.
Orientation: If a>b, main axis is horizontal, and the
vertices are the x-intercepts.
If b>a, main
axis is vertical, and the vertices are the y-intercepts.
Graph connects the four intercepts with an oval.
Ex: x2 / 4 + y2 / 16 = 1
Hyperbola centered at origin: x2 / a2 - y2 / b2 = 1 or y2 / b2 - x2 / a2 = 1
First form: x-intercepts: ±a Why? Set y=0, clear fraction.
No y-intercepts exist (you would be solving y2 = sqrt (-b2): no real solution.
Second form: y-intercepts: ±b (and there are no x-intercepts, because of the "-")
"Asymptotes" (lines which the graph approaches as |x| and/or |y| becomes large)
Asymptotes pass through the center (origin)
Slope of asymptotes is ±b/a (root of denom of y2-term / root of denom of x2-term)
To graph, plot the points (±a, ±b) and draw an X through the corners
Graph of hyperbola starts at each vertex (x- or y-intercept) and curves outward toward the asymptotes.
Ex: y2 / 1 - x2 / 9 = 1
Shifting: replacing x and y by (x-h) and (y-k) in the above equations shifts the center of the graph to (h,k);
It is useful to sketch shifted axes x' and y' through the new "origin" (h,k).
The shape of the graph does not change.
Therefore, the intercepts are at h ± a and y ± b; reference points for asymptotes are (h ± a, y ± b)
Ex: (x-2)2 / 9 - (y+1)2 / 4 =1: (h,k) = (2,-1), a=3, b=2; vertices at 2±3, asymptotes have slope ±2/3.
9.3 Second-Degree Inequalities and Nonlinear Systems
Second-Degree Inequalities
Graph, using appropriate (solid or broken) curve.
Pick test point to determine which region makes inequality true -- shade the true region.
Nonlinear Systems -- Solve by substitution (usually); Remember ± for square root principle.
You can substitute for x2 (or y2), rather than solving for x (or y), if that will clear a variable.
Return to: Merced College; Don Power Updated 11/03/08 by Don Power