8.13. Answer: The xintercept of a function y = f(x) is the intersection
the graph of f and the x axis. The xaxis is the graph of the
function g(x) = 0. Thus the xaxis 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.
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) = x^{2}+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) = 2x^{2}−5x+2 and g(x) = x+3.
These two curves intersect at
(b) Find points of intersection: f(x) = x^{2} + 4x − 1 and g(x) =
1 − 4x − x^{2}.

equate ordinates 

substitute 

add −1 + 4x + x^{2} 

pos. discrim. 

Quadratic formula 




These two curves intersect at
(c) Find points of intersection: f(x) = 3x^{2} + 1 and g(x) = x^{2} − 5x.

equate ordinates 

substitute 

add −1 + 4x + x^{2} 

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) = 2x^{2} + 3 and g(x) = x^{2} − 1.

equate ordinates 

substitute 

add −x^{2} + 1 both sides 
The equation x^{2} = −4 has no solutions; therefore, these two
curves do not intersect.
(c) f(x) = x^{2} + 2x + 2 and g(x) = x + 1
A negative discriminant (b^{2}−4ac < 0) implies that the equation
has not solution; therefore, these two equations do not intersect.
(d) f(x) = x^{2} − 2x − 4 and g(x) = 4x^{2} − 3

equate ordinates 

substitute 

add −4x^{2} + 3 

negative discriminant 
A negative discriminant implies the equation has no solution.
Solutions to Examples
8.1. Solution:
P : (x_{1}, y_{1}) = (−2, 4)
Q : (x_{2}, y_{2}) = (3,−1)
We take the difference in the first coordinates and the difference in
the second coordinates.
x_{1} − x_{2} = −2 −3 = −5
y_{1} − y_{2} = 4− (−1) = 4+1 = 5
We now take the sum of the squares of these two:
(x_{1} − x_{2})^{2} + (y_{1} − y_{2})^{2} = (−5)^{2} + 5^{2} = 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) = x^{2}+3x+1. The natural domain is
the set of all real numbers for which the value of f(x) = x^{2}+3x+1 can
be computed as a real number. For any real number x , the expression
x^{2} + 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  x^{2} − 3x + 2 ≠0}
This should not be considered to be a satis factory characterization of
the domain of g though.
First find where x^{2} − 3x + 2 = 0, and reason from there. 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 x^{2} − 1 ≥ 0 and x ≠ 0. Thus,
Dom(p) = { x  x^{2} − 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:
x^{2} − 1 ≥0 => x^{2} ≥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 lefthand 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 x^{2} − 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) = x^{2}−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: