Functions and Graphs - GRE Subject Test: Math

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Question

What is the equation of a circle with center at and a radius of ?

Answer

Step 1: Recall the general equation for a circle (if the vertex is not at :

, where center=

Step 2: Recall the shift of the graph..

If the value of is positive, it will be shown as a negative shift in the equation.
If the value of is negative, it will be shown as a positive shift in the equation.
If the value of is positive, it will be shown as a negative shift in the equation.
If the value of is negative, it will be shown as a positive shift in the equation.

Step 3: Look at the center given in the problem and find the rule(s) in step 2 that will apply:

Center=, ,

Step 4: Plug in into the equation of a circle:

Simplify:

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Question

What is the vertex of the equation of a circle:

Answer

Step 1: There are no numbers next to and , so their is no movement of the vertex..

Step 2: Recall the vertex of a circle that does not move...

The vertex of this circle is .

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Question

What kind of function is this: $f(x)=\sqrt[3] {x-1}$ ?

Answer

Step 1: Look at the equation.. $\sqrt[3] {x-1}$ . The cube-root outside of the function determines what the answer is..

The function is a cube-root function.

Note:

Square function,
Cube function,
Rational function, $\frac {1}{x}$ (if )

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Question

What kind of function is

Answer

Step 1: To classify any function, find the degree of that function. The degree of a function is the highest exponent.

Step 2: Find degree of .

The degree is 3..

Step 3: Find what kind of function has degree of 3:

A cubic function has a degree of 3...

A square function has a degree of 2
A line has a degree of 1

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Question

What kind of function is $\large y=x^2$

Answer

Step 1: Define some basic graphs and their functions...

Line functions have an equation $\large ax+b$ , where $\large \large {a,b}\in \mathbb{R}$
Parabola (Quadratic functions) have an equation $\large ax^2+bx+c$ or $\large ax^2+c$ or $\large ax^2$ , where $\large \large {a,b,c}\in \mathbb{R}$

Cubic functions have an equation $\large \large ax^3 \pm b$ , where $\large {a,b} \in \mathbb{R}$

Step 2: Determine what type of function is given...

The equation given is a parabola...

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Question

What kind of function is $\large y=\sqrt[3] {x+6}-4$ ?

Answer

Quadratic functions have at least an $\large x^2$ term.
Absolute Functions have $\large \left | x\right |$ .

Cube-Root functions have $\large \sqrt[3]{x}$

The function $\large y$ is a cube-root function because it shares $\large \sqrt[3]{x}$

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Question

Find the minimum distance between the point and the following line:

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

Answer

The minimum distance from the point to the line would be found by drawing a segment perpendicular to the line directly to the point. Our first step is to find the equation of the new line that connects the point to the line given in the problem. Because we know this new line is perpendicular to the line we're finding the distance to, we know its slope will be the negative inverse of the line its perpendicular to. So if the line we're finding the distance to is:

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

Then its slope is -1/3, so the slope of a line perpendicular to it would be 3. Now that we know the slope of the line that will give the shortest distance from the point to the given line, we can plug the coordinates of our point into the formula for a line to get the full equation of the new line:

$-4=3(-2)+b\rightarrow b=2$

Now that we know the equation of our perpendicular line, our next step is to find the point where it intersects the line given in the problem:

$3x+2=-\frac{1}{3}x+2$

$\frac{10}{3}x=0\rightarrow x=0$

So if the lines intersect at x=0, we plug that value into either equation to find the y coordinate of the point where the lines intersect, which is the point on the line closest to the point given in the problem and therefore tells us the location of the minimum distance from the point to the line:

So we now know we want to find the distance between the following two points:

and

Using the following formula for the distance between two points, which we can see is just an application of the Pythagorean Theorem, we can plug in the values of our two points and calculate the shortest distance between the point and line given in the problem:

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

$d=\sqrt{(0-(-2))^2+(2-(-4))^2}=\sqrt{(2)^2+(6)^2}=\sqrt{40}$

Which we can then simplify by factoring the radical:

$\sqrt{40}=\sqrt{4\cdot 10}=2\sqrt{10}$

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Question

What is the shortest distance between the line and the origin?

Answer

The shortest distance from a point to a line is always going to be along a path perpendicular to that line. To be perpendicular to our line, we need a slope of $-\frac{1}{3}$ .

To find the equation of our line, we can simply use point-slope form, using the origin, giving us

$y-0=\left(-\frac{1}{3}\right)(x-0)$ which simplifies to $y=-\frac{x}{3}$ .

Now we want to know where this line intersects with our given line. We simply set them equal to each other, giving us $-\frac{x}{3}=3x-4$ .

If we multiply each side by , we get .

We can then add to each side, giving us .

Finally we divide by , giving us $x=\frac{6}{5}$ .

This is the x-coordinate of their intersection. To find the y-coordinate, we plug $\frac{6}{5}$ into $y=-\frac{x}{3}$ , giving us $y=-\frac{2}{5}$ .

Therefore, our point of intersection must be $\left(\frac{6}{5},-\frac{2}{5}\right)$ .

We then use the distance formula $d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ using $\left(\frac{6}{5},-\frac{2}{5}\right)$ and the origin.

This give us $d=\sqrt{\left(\frac{36}{25}\right)+\left(\frac{4}{25}\right)}=\sqrt{\frac{40}{25}}=\frac{2\sqrt{10}}{5}$ .

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Question

Find the distance from point $\left ( 1,-\frac{1}{2}\right )$ to the line .

Answer

Draw a line that connects the point and intersects the line at a perpendicular angle.

The vertical distance from the point $\left ( 1,-\frac{1}{2}\right )$ to the line will be the difference of the 2 y-values.

The distance can never be negative.

$5-\left(-\frac{1}{2}\right) = \frac{10}{2}+\frac{1}{2}=\frac{11}{2}$

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Question

Find the distance between point $(\frac{1}{2},\frac{1}{2})$ to the line $x=-\frac{1}{2}$ .

Answer

Distance cannot be a negative number. The function $x=-\frac{1}{2}$ is a vertical line. Subtract the value of the line to the x-value of the given point to find the distance.

$D=\left | (-\frac{1}{2})-\frac{1}{2} \right | = \left | -1\right |=1$

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Question

Find the distance between point to line .

Answer

The line is vertical covering the first and fourth quadrant on the coordinate plane.

The x-value of is negative one.

Find the perpendicular distance from the point to the line by subtracting the values of the line and the x-value of the point.

Distance cannot be negative.

$\left | 1-(-1)\right |=\left |2 \right |=2$

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Question

Find the distance between the two lines.

$\ y_1=2.65x-13 \ y_2=2.65x+4$

Answer

Since the slope of the two lines are equivalent, we know that the lines are parallel. Therefore, they are separated by a constant distance. We can then find the distance between the two lines by using the formula for the distance from a point to a nonvertical line:

$d=\frac{|Ax_1 +By_1+C|}{\sqrt{A^{2}+B^{2}}}$

First, we need to take one of the line and convert it to standard form.

where $A=2.65,B=-1, \textup{and }C=-13$

Now we can substitute A, B, and C into our distance equation along with a point, $\left ( x_1,y_1\right )$ , from the other line. We can pick any point we want, as long as it is on line . Just plug in a number for x, and solve for y. I will use the y-intercept, where x = 0, because it is easy to calculate:

Now we have a point, , that is on the line . So let's plug our values for $x_1,y_1,A,B,\textup{and }C}$ :

$d=\frac{|(2.65)\cdot (0) +(-1)\cdot (4)+(-13)|}{\sqrt{(2.65)^{2}+(-1)^{2}}}=\frac{|0-4-13|}{\sqrt{(7.0225+1)}}=\frac{|-17|}{\sqrt{8.0225}}$

$=\frac{17}{2.83}=6.002\approx6$

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Question

How far apart are the line $y = \frac{ 1}{2 } x - 3$ and the point ?

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation $y = \frac{1}{2} x - 3$ in this form to identify a, b, and c:

$y = \frac{1}{2} x - 3$ subtract half x and add 3 to both sides

$y - \frac{1}{2}x + 3 = 0$ multiply both sides by 2

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|-1(2)+2(7)+6 |}{\sqrt{(-1)^2 +( 2)^2 }} = \frac{|-2 + 14+6|}{\sqrt{1+4} } = \frac{18}{\sqrt5}$

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Question

Find the distance between and

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation in this form to identify , , and :

subtract and from both sides

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|-4(2)+1(-3)-1 |}{\sqrt{(1)^2 +(-4)^2 }} = \frac{|-8 -3-1|}{\sqrt{1+16} } = \frac{12}{\sqrt{17}}$

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Question

How far apart are the line $y = -\frac{2}{3}x + 8$ and the point ?

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation $y = -\frac{2}{3} x +8$ in this form to identify a, b, and c:

$y = -\frac{2}{3} x+8$ add $\frac{2}{3}x$ to and subtract 8 from both sides

$y + \frac{2}{3}x -8 = 0$ multiply both sides by 3

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|2(1)+3(3)-24 |}{\sqrt{(3)^2 +( 2)^2 }} = \frac{|2 + 9-24|}{\sqrt{9+4} } = \frac{13}{\sqrt{13}}$

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Question

Find the distance between $y = -\frac{4}{5}x - 2$ and .

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation $y = -\frac{4}{5} x -2$ in this form to identify , , and :

$y = -\frac{4}{5} x-2$ add $\frac{4}{5}x$ and to both sides

$y + \frac{4}{5}x +2 = 0$ multiply both sides by

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|4(-3)+5(4)+10 |}{\sqrt{(4)^2 +(5)^2 }} = \frac{|-12+20+10|}{\sqrt{16+25} } = \frac{18}{\sqrt{41}}$

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Question

Find the distance between $y = \frac{7}{2} x - 14$ and

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation $y = \frac{7}{2} x -14$ in this form to identify , , and :

$y = \frac{7}{2} x-14$ subtract $\frac{7}{2}x$ from and add to both sides

$y -\frac{7}{2}x +14 = 0$ multiply both sides by

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|-7(-6)+2(2)+28 |}{\sqrt{(-7)^2 +( 2)^2 }} = \frac{|13 + 4+28|}{\sqrt{49+4} } = \frac{45}{\sqrt{53}}$

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Question

Find the distance between $y = \frac{12}{5} x - 3$ and the point

Answer

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is and the line is

First, we'll re-write the equation $y = \frac{12}{5} x-3$ in this form to identify , , and :

$y = \frac{12}{5} x-3$ subtract $\frac{12}{5}x$ from and add to both sides

$y -\frac{12}{5}x +3 = 0$ multiply both sides by

Now we see that . Plugging these plus into the formula, we get:

$distance = \frac{|-12(3)+5(-4)+15 |}{\sqrt{(5)^2 +( 12)^2 }} = \frac{|-36 + -20+15|}{\sqrt{25+144} } = \frac{41}{\sqrt{169}} = \frac{41}{13}$

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Question

Find the distance between and

Answer

To find the distance, choose any point on one of the lines. Plugging in 2 into the first equation can generate our first point:

this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

subtract the whole right side from both sides

now we see that

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where represents the point.

$\frac{|-2(2) + 1(1)-1| }{ \sqrt{1^2 + 2^2 }} = \frac{|-4+1-1|}{\sqrt{1+4}} = \frac{4}{\sqrt{5}}$

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Question

Find the distance between $y = \frac{1}5 x +4$ and $y = \frac{1}{5}x - 3$

Answer

To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = \frac{1}{5}(5) - 3 = 1 -3 = -2$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = \frac{1}{5}x+4$ subtract the whole right side from both sides

$- \frac{1}{5}x + y - 4 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where represents the point.

$\frac{|-1(5) + 5(-2)-20| }{ \sqrt{1^2 + 5^2 }} = \frac{|-5-10-20|}{\sqrt{1+25}} = \frac{35}{\sqrt{26}}$

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