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SAT II Math II : Geometry

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Example Questions

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Example Question #7 : Coordinate Geometry

Which of the following symmetries applies to the graph of the relation

$\left (x- 4 \right )^{2} + y ^{2}= 18$ ?

I) Symmetry with respect to the origin

II) Symmetry with respect to the -axis

III) Symmetry with respect to the -axis

Possible Answers:

I, II, and III

II only

I only

III only

None of these

Correct answer:

II only

Explanation:

The relation

$(x-h)^{2} + (y - k)^{2} = r^{2}$

is a circle with center (h,k) and radius .

In other words, it is a circle with center at the origin, translated right units and up units.

$\left (x- 4 \right )^{2} + y ^{2}= 18$

$\left (x- 4 \right )^{2} +\left ( y - 0 \right ) ^{2}= 18$

is a circle translated right 4 units and up zero units. The upshot is that the circle moves along the -axis only, and therefore is symmetric with respect to the -axis, but not the -axis. Also, as a consequence, it is not symmetric with respect to the origin.

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Example Question #1 : Coordinate Geometry

Axes_1

Refer to the above figure.

Which of the following functions is graphed?

Possible Answers:

g(x) = |x-3| + 6

g(x) = |x - 6| + 3

The correct answer is not given among the other responses.

g(x) = |x +6| + 3

g(x) = |x+3| + 6

Correct answer:

g(x) = |x - 6| + 3

Explanation:

Below is the graph of f(x) = |x| :

Axes_1

The given graph is the graph of shifted 6 units left (that is, unit right) and 3 units up.

The function graphed is therefore

g (x) = f(x-h) + k where h = 6, k= 3 . That is,

g (x) = f(x-6) +3

g(x)= | x-6 | + 3

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Example Question #2 : Coordinate Geometry

Axes_1

Refer to the above figure.

Which of the following functions is graphed?

Possible Answers:

g(x) = |3x +6|

g(x) = |2x| + 3

$g(x) = \left |3x + \frac{3}{2} \right |\textup{}$

$g(x) =\left | \frac{1}{2} x \right |+ 3$

g(x) = |2x + 6|

Correct answer:

g(x) = |2x| + 3

Explanation:

Below is the graph of f(x) = |x| :

Axes_1

If the graph of is translated by shifting each point (x,y) to the point (2, 2y) , the graph of

g(x) = 2f(x)

is formed. If the graph is then shifted upward by three units, the new graph is

g(x) = 2f(x) + 3

Since the starting graph was f(x) = |x| , the final graph is

g(x) = 2 |x| + 3 , or,

g(x) = |2x| + 3

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Example Question #1 : Coordinate Geometry

Axes_1

Refer to the above figure.

Which of the following functions is graphed?

Possible Answers:

g(x) =- |3x+12|

$g(x) =\left | \frac{1}{3} x+\frac{4}{3} \right |$

g(x) = |3x+12|

g(x) = |3x-12|

$g(x) =\left | \frac{1}{3} x-\frac{4}{3} \right |$

Correct answer:

g(x) = |3x-12|

Explanation:

Below is the graph of f(x) = |x| :

Axes_1

If the graph of is translated by shifting each point (x,y) to the point (x, 3y) , the graph of

g(x) = 3f(x)

is formed. If the graph is then shifted right by four units, the new graph is

g(x) = 3f(x- 4)

Since the starting graph was f(x) = |x| , the final graph is

g(x) = 3 |x-4| , or g(x) = |3x-12|

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Example Question #11 : Coordinate Geometry

Axes_1

Refer to the above figure.

Which of the following functions is graphed?

Possible Answers:

$g(x) =\left | \frac{1}{4}x \right |- 4$

$g(x) =\left | \frac{1}{4}x+1 \right |$

g(x) =-|4x |- 4

g(x) =-|4x+16|

$g(x) =\left | \frac{1}{4}x-1 \right |$

Correct answer:

$g(x) =\left | \frac{1}{4}x \right |- 4$

Explanation:

Below is the graph of f(x) = |x| :

Axes_1

If the graph of is translated by shifting each point (x,y) to the point $\left ( x, \frac{1}{4}y \right )$ , the graph of

$g(x) = \frac{1}{4}f(x)$

is formed. If the graph is then shifted down by four units, the new graph is

$g(x) = \frac{1}{4}f(x)- 4$ .

Since the starting graph was f(x) = |x| , the final graph is

$g(x) = \frac{1}{4}|x|- 4$ , or

$g(x) =\left | \frac{1}{4}x \right |- 4$

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Example Question #12 : Coordinate Geometry

Axes_1

Refer to the above figure.

Which of the following functions is graphed?

Possible Answers:

g(x) = |3x-6|

$g(x) =\left | \frac{1}{3} x \right |+2$

g(x) = |3x|-2

$g(x) =\left | \frac{1}{3} x+ \frac{2}{3} \right |$

$g(x) =\left | \frac{1}{3} x- \frac{2}{3} \right |$

Correct answer:

$g(x) =\left | \frac{1}{3} x- \frac{2}{3} \right |$

Explanation:

Below is the graph of f(x) = |x| :

Axes_1

If the graph of is translated by shifting each point (x,y) to the point $\left ( x, \frac{1}{3}y \right )$ , the graph of

$g(x) = \frac{1}{3}f(x)$

is formed. If the graph is then shifted right by two units, the new graph is

$g(x) = \frac{1}{3}f(x-2)$

Since the starting graph was f(x) = |x| , the final graph is

$g(x) = \frac{1}{3} |x-2|$ , or

$g(x) =\left | \frac{1}{3} x- \frac{2}{3} \right |$

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Example Question #1 : Distance Formula

What is the distance between the point (1,2) and (8,5)?

Possible Answers:

13.9

$\sqrt58$

Correct answer:

$\sqrt58$

Explanation:

For this question we will use the distance formula to solve.

$d=\sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

In our case (x_1, y_1)=(1,2)

and (x_2, y_2)= (8,5)

Substituting these values in we get the following

$d=\sqrt((8-1)^2+(5-2)^2)$

$d=\sqrt(7^2+3^2)$

$d=\sqrt(49+9)$

$d=\sqrt(58)$

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Example Question #1 : Distance Formula

The points A=(-2,0), B=(0,3), and C=(0,0) makes a triangle. What is the distance between point A and point B?

Possible Answers:

$\sqrt15$

$\sqrt13$

Correct answer:

$\sqrt13$

Explanation:

For this question we need to use the distance formula for points A and B.

Point A will be our (x_1, y_1) and point B will be (x_2, y_2)

Now we substitute these values into the following:

$d=\sqrt((0--2)^2+(3-0)^2)$

$d=\sqrt(2^2+3^2)$

$d=\sqrt(4+9)$

$d=\sqrt13$

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Example Question #11 : Coordinate Geometry

Find the distance between the two points (2,7) and (4,6).

Possible Answers:

$\sqrt{17}$

$\sqrt{3}$

$\sqrt{5}$

$\sqrt{205}$

Correct answer:

$\sqrt{5}$

Explanation:

The distance between two points is found using the formula $d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

For this problem the values are as follows:

$x_{1}=2$

$x_{2}=4$

$y_{1}=7$

$y_{2}=6$

Input the values into the formula and simplify

$d=\sqrt{(4-2)^2+(7-6)^2}=\sqrt{(2)^2+(1)^2}=\sqrt{4+1}=\sqrt{5}$

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Example Question #16 : Coordinate Geometry

Find the point halfway between points A and B.

A=(19,54)

B=(67,34)

Possible Answers:

(86,88)

(43,44)

(42,45)

(44,43)

Correct answer:

(43,44)

Explanation:

Find the point halfway between points A and B.

A=(19,54)

B=(67,34)

We are going to need to use midpoint formula. If you ever have difficulty recalling midpoint formula, try to recall that it is basically taking two averages. One average is the average of your x values, the other average is the average of your y values.

$Mid=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Now we plug and chug!

$Mid=(\frac{19+67}{2},\frac{54+34}{2})=(\frac{86}{2},\frac{88}{2})=(43,44)$

So our answer is (43,44)

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SAT II Math II : Geometry

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #7 : Coordinate Geometry

Example Question #1 : Coordinate Geometry

Example Question #2 : Coordinate Geometry

Example Question #1 : Coordinate Geometry

Example Question #11 : Coordinate Geometry

Example Question #12 : Coordinate Geometry

Example Question #1 : Distance Formula

Example Question #1 : Distance Formula

Example Question #11 : Coordinate Geometry

Example Question #16 : Coordinate Geometry

All SAT II Math II Resources

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