Geometry - Graphing | Practice Hub

Give the domain of the function

$f(x)= x^{2}-3x-10$

The set of all real numbers

$\left { x| x <-5 \textrm{ or } x >2 \right }$

$\left { x| x \le -5 \textrm{ or } x \ge 2 \right }$

$\left { x| -5 < x < 2 \right }$

$\left { x| -5 \le x \le 2 \right }$

Explanation

The domain of any polynomial function, such as , is the set of real numbers, as a polynomial can be evaluated for any real value of .

$f(x)= Ax^{2}+ Bx+ C$ has as its graph a vertical parabola on the coordinate plane. You are given that and , but you are not given .

Which of the following can you determine without knowing the value of ?

I) Whether the graph is concave upward or concave downward

II) The location of the vertex

III) The location of the -intercept

IV) The locations of the -intercepts, if there are any

V) The equation of the line of symmetry

I and III only

I and V only

I, II, and V only

I, III, and IV only

III and IV only

Explanation

I) The orientation of the parabola is determined solely by the sign of . Since , the parabola can be determined to be concave downward.

II and V) The -coordinate of the vertex is $-\frac{B}{2A}$ ; since you are not given , you cannot find this. Also, since the line of symmetry has equation $x=-\frac{B}{2A}$ , for the same reason, you cannot find this either.

III) The -intercept is the point at which ; by substitution, it can be found to be at . known to be equal to 9, so the -intercept can be determined to be .

IV) The -intercept(s), if any, are the point(s) at which $f(x)= Ax^{2}+ Bx+ C = 0$ . This is solvable using the quadratic formula

$x= \frac{-B \pm \sqrt{B^{2}-4AC}}{2A}$

Since all three of and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I and III only.

The point on the coordinate plane with coordinates $\left ( 4 \frac{1}{2}, 6 \frac{1}{3}\right )$ lies _____

in Quadrant I.

in Quadrant II.

in Quadrant III.

in Quadrant IV.

on an axis.

Explanation

On the coordinate plane, a point with a positive -coordinate and a positive -coordinate lies in the upper right quadrant - Quadrant I.

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.

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.

Define $g (x) = \frac{1}{\sqrt[3]{x -27}}$

What is the natural domain of ?

$(-\infty , 27)\cup (27, \infty )$

$(27, \infty )$

$(3, \infty )$

$(-\infty , - 27)\cup (27, \infty )$

$(-\infty , - 3)\cup (3, \infty )$

Explanation

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression $\sqrt[3]{x -27}$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which