Advanced Geometry

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Questions 1 - 10

Define a function as follows:

$f(x) = 2 ^{x-1} + 7$

Give the horizontal aysmptote of the graph of .

$y = -\frac{7}{2}$

Explanation

The horizontal asymptote of an exponential function can be found by noting that a positive number raised to any power must be positive. Therefore, $2 ^{x-1} > 0$ and $y= 2 ^{x-1} + 7 > 7$ for all real values of . The graph will never crosst the line of the equatin , so this is the horizontal asymptote.

What is the domain of $y = 4 - x^{2}$ ?

all real numbers

$x \leq 4$

$x \geq 4$

$x \leq 0$

Explanation

The domain of the function specifies the values that can take. Here, $4-x^{2}$ is defined for every value of , so the domain is all real numbers.

Give the -coordinate(s) of the -intercept(s) of the graph of the function

$f(x) = \frac{1}{2} x- \frac{7}{2}$

$\frac{1}{2}$

$- \frac{7}{2}$

$\frac{1}{7}$

The graph of has no -intercept.

Explanation

The -intercept(s) of the graph of are the point(s) at which it intersects the -axis. The -coordinate of each is 0; their -coordinate(s) are those value(s) of for which , so set up, and solve for , the equation:

$\frac{1}{2} x- \frac{7}{2} =0$

Add $\frac{7}{2}$ to both sides:

$\frac{1}{2} x- \frac{7}{2} + \frac{7}{2} =0+ \frac{7}{2}$

$\frac{1}{2}x =\frac{7}{2}$

Multiply both sides by 2:

$2 \cdot \frac{1}{2}x =2 \cdot \frac{7}{2}$

the correct choice.

Two congruent equilateral triangles with sides of length are connected so that they share a side. Each triangle has a height of . Express the area of the shape in terms of .

$\frac{h}{2}$

Explanation

The shape being described is a rhombus with side lengths 1. Since they are equilateral triangles connected by one side, that side becomes the lesser diagonal, so .

The greater diagonal is twice the height of the equaliteral triangles, .

The area of a rhombus is half the product of the diagonals, so:

$A=\frac{pq}{2}=\frac{1 \cdot 2h}{2}=h$

Give the domain of the function

$f(x) = \sqrt{x^{2}+ 25}$

The set of all real numbers

$\left \{ x | x \ge 5 \right \}$

$\left \{ x | x \ge -5 \right \}$

$\left \{ x | x \ge -25 \right \}$

$\left \{ x | x \ge 25 \right \}$

Explanation

The function $f(x) = \sqrt{x^{2}+ 25}$ is defined for those values of for which the radicand is nonnegative - that is, for which

$x^{2}+ 25 \ge 0$

Subtract 25 from both sides:

$x^{2}+ 25- 25 \ge 0 - 25$

$x^{2} \ge - 25$

Since the square root of a real number is always nonnegative,

$x^{2} \ge 0 > - 25$

for all real numbers . Since the radicand is always positive, this makes the domain of the set of all real numbers.

Which of the following shapes is a trapezoid?

Shapes

Explanation

A trapezoid is a four-sided shape with straight sides that has a pair of opposite parallel sides. The other sides may or may not be parallel. A square and a rectangle are both considered trapezoids.

Give the domain of the function

$f(x) =x+ \sqrt{x+ 7}$

$\left \{ x| x \ge -7 \right \}$

The set of all real numbers

$\left \{ x| x \ne -7 \right \}$

$\left \{ x| x \ne 7 \right \}$

$\left \{ x| x \ge 7 \right \}$

Explanation

The square root of a real number is defined only for nonnegative radicands; therefore, the domain of is exactly those values for which the radicand is nonnegative. Solve the inequality:

$x+ 7 \ge 0$

$x+ 7 - 7 \ge 0 - 7$

$x \ge - 7$

The domain of is $\left \{ x| x \ge -7 \right \}$ .

Define a function as follows:

$f(x) = 2 ^{x-3}+ 5$

Give the vertical aysmptote of the graph of .

The graph of does not have a vertical asymptote.

$x= - \frac{5}{2}$

Explanation

Since any number, positive or negative, can appear as an exponent, the domain of the function $f(x) = 2 ^{x-3}+ 5$ is the set of all real numbers; in other words, is defined for all real values of . It is therefore impossible for the graph to have a vertical asymptote.