Geometry - Coordinate Geometry

$Find\ the\ vertex\ for\ the\ function: f(x)=4x^{2}-16x+11.$

Explanation

$Use \frac{-b}{2a}\ to\ calculate\ the\ x-value\ of\ the\ vertex.$

$\small Given\ the\ function\ f(x)=4x^{2}-16x+11\ is\ in\ standard\ form\ y=ax^{2}+bx+c$

$\small We\ can\ see\ that\ a=4, b=-16\ and\ c=11$

$\small Substitute\ these\ values\ in\ for\ x=\frac{-b}{2a}\ to\ get\ x=\frac{16}{8}=2$

$\small Now\ substitute\ x=2\ into\ the\ function\ f(x)\ to\ get:$

$\small f(2)=4(2)^{2}-16(2)+11=-5$

$\small The\ coordinates\ of\ the\ vertex\ are\ (2,-5)$

$f(x)= Ax^{2}+ Bx+ C$ 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.

Give the domain of the function

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

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

The set of all real numbers

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

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

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

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.

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.

In which quadrant does the complex number lie?

Explanation

If we graphed the given complex number on a set of real-imaginary axes, we would plot the real value of the complex number as the x coordinate, and the imaginary value of the complex number as the y coordinate. Because the given complex number is as follows:

We are essentially doing the same as plotting the point on a set of Cartesian axes. We move units right in the x direction, and units down in the y direction, which puts us in the fourth quadrant, or in terms of Roman numerals:

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.

$Find\ the\ vertex\ for\ the\ function: f(x)=4x^{2}-16x+11.$

Explanation

$Use \frac{-b}{2a}\ to\ calculate\ the\ x-value\ of\ the\ vertex.$

$\small Given\ the\ function\ f(x)=4x^{2}-16x+11\ is\ in\ standard\ form\ y=ax^{2}+bx+c$

$\small We\ can\ see\ that\ a=4, b=-16\ and\ c=11$

$\small Substitute\ these\ values\ in\ for\ x=\frac{-b}{2a}\ to\ get\ x=\frac{16}{8}=2$

$\small Now\ substitute\ x=2\ into\ the\ function\ f(x)\ to\ get:$

$\small f(2)=4(2)^{2}-16(2)+11=-5$

$\small The\ coordinates\ of\ the\ vertex\ are\ (2,-5)$

$f(x)= Ax^{2}+ Bx+ C$ 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.

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 $\left ( 100, 0 \right )$ lies _____

on an axis.

in Quadrant IV.

in Quadrant III.

in Quadrant I.

in Quadrant II.

Explanation

On the coordinate plane, a point with 0 as one of its coordinates lies on an axis.

Coordinate Geometry

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