College Algebra Exercises Section 3.1

As you study these exercises, move your cursor over the light bulbs and the highlighted words.
Study with an active cursor

The definition that pops up when you move the cursor over a highlighted word is what should pop into your mind when you read, hear, or speak that word. The material that pops up when you move the cursor over a light bulb is a suggested strategy for solving the problem. You should always formulate a similar strategy when you attempt to solve a problem. When you move the cursor over an explanation of a step in a solution, the property that pops up is the mathematical justification for taking that action. You must always be able to provide such a justification for every step you take in mathematics.

1 & 2 )          

The problem merely requests that you match function with graph . However, seeing these two graphs should prompt you to ask some additional questions.
a) What are the y-intercepts of the two graphs?
b) Do the two graphs intersect? If so, where do they intersect?
c) Does the formula   yield the coordinates of the points which appear to be the vertex on each of the graphs?

3 & 4 )          

The problem merely requests that you match function with graph . However, seeing these two graphs should prompt you to ask some additional questions.
a) What are the x-intercepts of the two graphs?
b) Do the two graphs intersect? If so, where do they intersect?
c) Does the formula   yield the coordinates of the points which appear to be the vertex on each of the graphs?

5 & 6 )          

The problem merely requests that you match function with graph . However, seeing these two graphs should prompt you to ask some additional questions.
a) What are the x-intercepts of the two graphs?
b) What are the y-intercepts of the two graphs?
c) Do the two graphs intersect? If so, where do they intersect?
d) Does the formula   yield the coordinates of the points which appear to be the vertex on each of the graphs?

14) Consider the function h whose rule is h(x) = 25 - x2
Discussion:

The function h is a quadratic function with negative leading coefficient. Therefore its graph is a parabola which opens down.  
The domain of h is the set of real numbers.
If x = 0, then h(x) = 25. Therefore the y-intercept is (0, 25)
The zeros of h are found by solving the equation resulting from h(x) = 0.
So we solve 0 = h(x) = 25 - x2 to find solutions 5 and -5.   These zeros are both real numbers and are therefore x-intercepts.
The vertex is
= (0, h(0)) = (0, 25)
An examination of the graph of h (shown at the right) reveals that:
h(x) > 0 for x {x | -5 < x < 5 } and
h(x) < 0 for x {x | x < -5 or x > 5 }

18) Consider the function f whose rule is f(x) = (x - 6)2 + 3 = x2 - 12x + 39
Discussion:

The function f is a quadratic function with positive leading coefficient. Therefore its graph is a parabola which opens up.
The domain of f is all real numbers.
If x = 0, then f(x) = 39. Therefore the y-intercept is (0, 39)
The zeros of f are found by solving the equation resulting from f(x) = 0.
So we solve 0 = f(x) = x2 - 12x + 39 with the quadratic formula.   The solutions are
These zeros are complex numbers and therefore are not x-intercepts. Therefore the graph of f does not intersect the x-axis.
The vertex is
= (6, f(6)) = (6, 3)
Examination of the graph of f shows that f(x) > 0 for all real numbers.

20) Consider the function f whose rule is f(x) = x2 + 2x + 1
Discussion:

The function f is a quadratic function with positive leading coefficient. Therefore its graph is a parabola which opens up.
The domain of f is all real numbers.
If x = 0, then f(x) = 1. Therefore the y-intercept is (0, 1)
The zeros of f are found by solving the equation resulting from f(x) = 0.
So we solve 0 = f(x) = x2 + 2x + 1 = (x + 1)(x + 1) = (x + 1)2.
The solution is -1.
This zero is a real number and therefore it is an x-intercept.
The zero has multiplicity 2 so the graph intersects but does not cross the axis at the zero.
Note that in the case of a parabola this permits the conclusion that the vertex of the parabola is the x-intercept. This will be verified with the following computation.
The vertex is
= (-1, f(-1)) = (-1, 0)
Examination of the graph of f shows that f(x) > 0 for all real numbers except -1.