Your Algebra Homework Can Now Be Easier Than Ever!

Cartesian Coordinate System and Functions

8.13. Answer: The x-intercept of a function y = f(x) is the intersection
the graph of f and the x -axis. The x-axis is the graph of the
function g(x) = 0. Thus the x-axis is the intersection of two curves :
y = f(x) and g(x) = 0.

8.14. Solutions: Hopefully, you used standard procedures.
(a) Find points of intersection: f(x) = 6x + 3 and g(x) = 2x − 7.

equate ordinates
substitute
add −2x − 3 both sides
divide by 4
d one !


When
Presentation of Answer :

(b) Find points of intersection: f(x) = x + 3 and g(x) = 2 − 8x.

equate ordinates
substitute
add 8x − 3 both sides
divide by 9


When
Presentation of Answer:

(c) Find points of intersection: f(x) = x2+7x−1 and g(x) = 4x−3.

equate ordinates
substitute
add −4x + 3 both sides
factor
done!


Calculation of Ordinates :
When x = −1, y = g(2) = −7.
When x = −2, y = g(3) = −11.
Presentation of Answer:

8.15. Solutions: We use standard procedures around here . . . how
about you?
(a) Find points of intersection: f(x) = 2x2−5x+2 and g(x) = x+3.

equate ordinates
substitute
add −x − 3 both sides
pos. discrim.
Quadratic formula
 


These two curves intersect at

(b) Find points of intersection: f(x) = x2 + 4x − 1 and g(x) =
1 − 4x − x2.

equate ordinates
substitute
add −1 + 4x + x2
pos. discrim.
Quadratic formula
 
 


These two curves intersect at

(c) Find points of intersection: f(x) = 3x2 + 1 and g(x) = x2 − 5x.

equate ordinates
substitute
add −1 + 4x + x2
pos. discrim.
Quadratic formula


These two curves intersect at

8.16. Solutions: Fol low the procedure.
(a) f(x) = 4x − 2 and g(x) = 4x + 12

equate ordinates
substitute
add −4x + 2 both sides


The equation 0 = 14 has no solution; i.e., no value of x can
satisfythe equation 0 = 14. Therefore, these two curves do not
intersect.

b) f(x) = 2x2 + 3 and g(x) = x2 − 1.

equate ordinates
substitute
add −x2 + 1 both sides


The equation x2 = −4 has no solutions; therefore, these two
curves do not intersect.

(c) f(x) = x2 + 2x + 2 and g(x) = x + 1

equate ordinates
substitute
add −x − 3 both sides
negative discriminant


A negative discriminant (b2−4ac < 0) implies that the equation
has not solution; therefore, these two equations do not intersect.

(d) f(x) = x2 − 2x − 4 and g(x) = 4x2 − 3

equate ordinates
substitute
add −4x2 + 3
negative discriminant


A negative discriminant implies the equation has no solution.

Solutions to Examples

8.1. Solution:
P : (x1, y1) = (−2, 4)
Q : (x2, y2) = (3,−1)
We take the difference in the first coordinates and the difference in
the second coordinates.
x1 − x2 = −2 −3 = −5
y1 − y2 = 4− (−1) = 4+1 = 5
We now take the sum of the squares of these two:
(x1 − x2)2 + (y1 − y2)2 = (−5)2 + 52 = 25+25 = 50.

Finally, we take the square root of this result:

Presentation of Solution:
Of course, this process can be accelerated once you fully understand
the computational steps.

8.2. Solution to: (a) Define f(x) = x2+3x+1. The natural domain is
the set of all real numbers for which the value of f(x) = x2+3x+1 can
be computed as a real number. For any real number x , the expression
x2 + 3x + 1 evaluates to a real number. Therefore, we deduce,

Solution to: (b) Define The numerator and denominator
always evaluate to a real number; however, if the denominator
evaluates to zero, the quotient is not a real number. Thus, we
can saythat
Dom(g) = { x | x2 − 3x + 2 ≠0}

This should not be considered to be a satis factory characterization of
the domain of g though.

First find where x2 − 3x + 2 = 0, and reason from there. Solve

given
factor
solve


Therefore,

 
set notation
interval notation


Solution to: (c) Define For any x, x + 2 evaluates to
a real number, but for to evaluate to a real number we must
have x + 2 ≥ 0. Thus,
Dom(h) = { x | x + 2 ≥ 0 }

Again, we should not be satisfied with this formulation. We next solve
the inequality:
x + 2 ≥0 => x ≥ −2

Thus,

Solution to: (d) Define In order for p(x) to evaluate
to a real number, we require x2 − 1 ≥ 0 and x ≠ 0. Thus,
Dom(p) = { x | x2 − 1 ≥ 0 }

We need to solve the inequality x 2 − 1 ≥ 0. To do this, we use the
Sign Chart Method originallydiscussed in Lesson 7. (Actually, this
method reallyisn’t needed for this simple inequality . We could solve
as follows:
x2 − 1 ≥0 => x2 ≥1 => |x| ≥ 1,
but we shall use the Sign Chatr Method in any case, just to remind
you of this method.)

We begin by factoring completely the left-hand side (which is a difference
of squares):
(x + 1)(x − 1) ≥ 0
The Sign Chart of (x + 1)(x − 1)

Therefore, the solution to the inequality x2 − 1 ≥ 0 is
(−∞,−1 ] ∪ [ 1,+∞)
But the solution to this inequalityis the natural domain of p. Thus,

Notice how all the techniques of algebra (Lessons 1–7) are used:
factoring, solving inequalities, interval notation and so on.

This is the discouraging and challanging thing about mathematics:
To solve any given problem, we must call on our entire history of
experiences in mathematics. This is why it is so important for us to
try to master each of the little steps we take toward our final goals.

8.3. Solution: Consider Based on the above strategy,
we see that

The first condition, x ≠ −1, avoids having zero in the denominator (we
exclude x = −1 from the domain); the second condition is necessary
for the radicand to be nonnegative. (The square root of a nonnegative
number is a real number, whereas the square root of a negative number
is a complex number . We don’t want to work with complex numbers
at this time.)

As you can see, I’ve simply translated the strategy into a series of
inequalities. We solve the inequality

first using the Sign Chart Methods.

The Sign Chart of

Therefore,

But, we also have the condition x ≠ −1; this changes the above solution
slightly to
x < −1 or x ≥ 0.

Presentation of Answer:

8.4. Solutions: We follow the standard procedures.
(a) Find the points of intersection of f(x) = 3x+2 and g(x) = 5x−4.

equate ordinates
substitute
add −5x − 2 to both sides
divide by 3


At x = 3, f(3) = 3(3) + 2 = 11. Thus, ( 3, 11 ) is the point of
intersection.
Presentation of Solution: Intersection Point(s):

(b) Find intersection points of f(x) = x2−3x+1 and g(x) = 2x−5.

equate ordinates
substitute
add −2x + 5 both sides
factor
done!


Calculation of Ordinates:
When x = 2, y = g(2) = −1.
When x = 3, y = g(3) = 1.

Presentation of Answer:
Points of Intersection:

Prev Next

Start solving your Algebra Problems in next 5 minutes!

Algebra Helper
Download (and optional CD)

Only $39.99

Click to Buy Now:


OR

2Checkout.com is an authorized reseller
of goods provided by Sofmath

Attention: We are currently running a special promotional offer for Algebra-Answer.com visitors -- if you order Algebra Helper by midnight of December 14th you will pay only $39.99 instead of our regular price of $74.99 -- this is $35 in savings ! In order to take advantage of this offer, you need to order by clicking on one of the buttons on the left, not through our regular order page.

If you order now you will also receive 30 minute live session from tutor.com for a 1$!

You Will Learn Algebra Better - Guaranteed!

Just take a look how incredibly simple Algebra Helper is:

Step 1 : Enter your homework problem in an easy WYSIWYG (What you see is what you get) algebra editor:

Step 2 : Let Algebra Helper solve it:

Step 3 : Ask for an explanation for the steps you don't understand:



Algebra Helper can solve problems in all the following areas:

  • simplification of algebraic expressions (operations with polynomials (simplifying, degree, synthetic division...), exponential expressions, fractions and roots (radicals), absolute values)
  • factoring and expanding expressions
  • finding LCM and GCF
  • (simplifying, rationalizing complex denominators...)
  • solving linear, quadratic and many other equations and inequalities (including basic logarithmic and exponential equations)
  • solving a system of two and three linear equations (including Cramer's rule)
  • graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions)
  • graphing general functions
  • operations with functions (composition, inverse, range, domain...)
  • simplifying logarithms
  • basic geometry and trigonometry (similarity, calculating trig functions, right triangle...)
  • arithmetic and other pre-algebra topics (ratios, proportions, measurements...)

ORDER NOW!

Algebra Helper
Download (and optional CD)

Only $39.99

Click to Buy Now:


OR

2Checkout.com is an authorized reseller
of goods provided by Sofmath
Check out our demo!
 
"It really helped me with my homework.  I was stuck on some problems and your software walked me step by step through the process..."
C. Sievert, KY
 
 
Sofmath
19179 Blanco #105-234
San Antonio, TX 78258
Phone: (512) 788-5675
Fax: (512) 519-1805
 

Home   : :   Features   : :   Demo   : :   FAQ   : :   Order

Copyright © 2004-2017, Algebra-Answer.Com.  All rights reserved.