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Precalculus : Graphing Functions

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Example Questions

Example Question #91 : Graphing Functions

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

Possible Answers:

$\frac{7}{\sqrt{29}}$

$\frac{5}{\sqrt{29}}$

$\frac{35}{\sqrt{26}}$

$\frac{7}{\sqrt{26}}$

$\frac{5}{\sqrt{26}}$

Correct answer:

$\frac{35}{\sqrt{26}}$

Explanation:

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 (5, -2)

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

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

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

-x + 5y-20=0

now we see that a = -1, b = 5, c = -20

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where (x,y) 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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Example Question #11 : Distance

How far apart are the lines $y = -\frac{3}{2}x -3$ and $y = -\frac{3}{2}x + 6$ ?

Possible Answers:

$\frac{6}{\sqrt{13}}$

$\frac{26}{\sqrt{13}}$

$\frac{18}{\sqrt{13}}$

$\frac{30}{\sqrt{40}}$

$\frac{30}{\sqrt{13}}$

Correct answer:

$\frac{18}{\sqrt{13}}$

Explanation:

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

$y = -\frac{3}{2}(2) - 3 = -3 - 3 =-6$ this gives us the point (2, -6)

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

$y =- \frac{3}{2}x+6$ subtract the whole right side from both sides

$\frac{3}{2}x + y - 6 = 0$ multiply both sides by

3x +2y - 12 = 0

now we see that a = 2, b = 3, c = -12

We can plug the coefficients and the point into the formula

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

$\frac{|3(2) + 2(-6)-12| }{ \sqrt{3^2 + 2^2 }} = \frac{|6-12-12|}{\sqrt{9+4}} = \frac{18}{\sqrt{13}}$

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Example Question #31 : Calculus

Find the distance between $y = \frac{4}{3}x +12$ and $y = \frac{4}{3}x + 2$

Possible Answers:

$\frac{51}{\sqrt{45}}$

$\frac{30}{\sqrt{45}}$

$\frac{51}{5}$

Correct answer:

Explanation:

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

$y = \frac{4}{3}(3) + 2 = 4 + 2 = 6$ this gives us the point (3,6)

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

$y = \frac{4}{3}x + 12$ subtract the whole right side from both sides

$- \frac{4}{3}x + y - 12 = 0$ multiply both sides by

-4x + 3 y -36 = 0

now we see that a = -4, b = 3, c = -36

We can plug the coefficients and the point into the formula

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

$\frac{|-4(3) +3(6)-36| }{ \sqrt{(-4)^2 + 3^2 }} = \frac{|-12+18-36|}{\sqrt{16+9}} = \frac{30}{\sqrt{25}} = \frac{30}{5} = 6$

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Example Question #2 : Find The Distance Between Two Parallel Lines

Find the distance between $y = -\frac{1}{2}x + 7$ and $y = -\frac{1}{2}x + 4$

Possible Answers:

$\frac{42}{\sqrt{34}}$

$\frac{18}{\sqrt{34}}$

$\frac{2}{\sqrt{5}}$

$\frac{9}{\sqrt{34}}$

$\frac{9}{\sqrt{5}}$

Correct answer:

$\frac{2}{\sqrt{5}}$

Explanation:

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

$y = -\frac{1}{2}(4)+7 = -2 + 7 = 5$ this gives us the point (4,5)

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

$y =-\frac{1}{2}x + 4$ subtract the whole right side from both sides

$- \frac{1}{2}x + y - 4 = 0$ multiply both sides by

x -2y + 8 = 0

now we see that a = 1, b = -2, c =8

We can plug the coefficients and the point into the formula

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

$\frac{|1(4) + -2(5)+8| }{ \sqrt{1^2 + (-2)^2 }} = \frac{|4-10+8|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$

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Example Question #31 : Gre Subject Test: Math

Find the distance between the lines $y = \frac{1}{2} x - 11$ and $y = \frac{1}{2} x + 1$

Possible Answers:

$\frac{24}{\sqrt5}$

$\frac{30}{\sqrt5}$

$\frac{24}{\sqrt{52}}$

Correct answer:

$\frac{24}{\sqrt5}$

Explanation:

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

$y = \frac{1}{2}(6) +1 = 3+1= 4$ this gives us the point (6,4)

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

$y = \frac{1}{2} x - 11$ subtract the whole right side from both sides

$-\frac{1}{2} x + y + 11 = 0$ multiply both sides by

-x + 2y + 22 = 0 now we see that a = -1, b = 2, c = 22

We can plug the coefficients and the point into the formula

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

$\frac{|-1(6) + 2(4)+22| }{ \sqrt{(-1)^2 + 2^2 }} = \frac{|-6+8+22|}{\sqrt{1+4}} = \frac{24}{\sqrt{5}}$

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Example Question #1 : Find Roots Of Quadratic Equation Using Discriminant

True or false: for a quadratic function of form ax²+ bx + c = 0, if the discriminant b²- 4ac = 0, there is exactly one real root.

Possible Answers:

True

False

Correct answer:

True

Explanation:

This is true. The discriminant b²- 4ac is the part of the quadratic formula that lives inside of a square root function. As you plug in the constants a, b, and c into b²- 4ac and evaluate, three cases can happen:

b²- 4ac > 0

b²- 4ac = 0

b²- 4ac < 0

In the first case, having a positive number under a square root function will yield a result that is a positive number answer. However, because the quadratic function includes $\pm$ , this scenario yields two real results.

In the middle case (the case of our example), $\sqrt0 = 0$ . Going back to the quadratic formula $x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$ , you can see that when everything under the square root is simply 0, then you get only $x = \frac{-b }{2a}$ , which is why you have exactly one real root.

For the final case, if b²- 4ac < 0, that means you have a negative number under a square root. This means that you will not have any real roots of the equation; however, you will have exactly two imaginary roots of the equation.

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Example Question #2 : Find Roots Of Quadratic Equation Using Discriminant

True or false: for a quadratic function of form ax²+ bx + c = 0, if the discriminant b²- 4ac > 0, there are exactly 2 distinct real roots of the equation.

Possible Answers:

True

False

Correct answer:

True

Explanation:

b²- 4ac > 0

b²- 4ac = 0

b²- 4ac < 0

In the first case (the case of our example), having a positive number under a square root function will yield a result that is a positive number answer. However, because the quadratic function includes $\pm$ , this scenario yields two real results.

In the middle case, $\sqrt0 = 0$ . Going back to the quadratic formula $x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$ , you can see that when everything under the square root is simply 0, then you get only $x = \frac{-b }{2a}$ , which is why you have exactly one real root.

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Example Question #3 : Find Roots Of Quadratic Equation Using Discriminant

True or false: for a quadratic function of form ax²+ bx + c = 0, if the discriminant b²- 4ac < 0, there are exactly two distinct real roots.

Possible Answers:

True

False

Correct answer:

False

Explanation:

This is false. The discriminant b²- 4ac is the part of the quadratic formula that lives inside of a square root function. As you plug in the constants a, b, and c into b²- 4ac and evaluate, three cases can happen:

b²- 4ac > 0

b²- 4ac = 0

b²- 4ac < 0

For the final case (the case of our example), if b²- 4ac < 0, that means you have a negative number under a square root. This means that you will not have any real roots of the equation; however, you will have exactly two imaginary roots of the equation.

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Example Question #1 : Find Roots Of Quadratic Equation Using Discriminant

Use the formula b² - 4ac to find the discriminant of the following equation: 4x² + 19x - 5 = 0.

Then state how many roots it has, and whether they are real or imaginary. Finally, use the quadratic function to find the exact roots of the equation.

Possible Answers:

Discriminant: 0

One real root: $\frac{-19}{8}$

Discriminant: 281

Two imaginary roots: $x = \frac{19\pm 281i}{8}$

Discriminant: 441

Two real roots: $x=\frac{1}{4}$ or x=-5

Discriminant: 441

Two real roots: $x=-\frac{1}{4}$ or x=5

Discriminant: 281

Two imaginary roots: $\frac{-19\pm 281i}{8}$

Correct answer:

Discriminant: 441

Two real roots: $x=\frac{1}{4}$ or x=-5

Explanation:

In the above equation, a = 4, b = 19, and c = -5. Therefore:

b² - 4ac = (19)² - 4(4)(-5) = 361 + 80 = 441.

When the discriminant is greater than 0, there are two distinct real roots. When the discriminant is equal to 0, there is exactly one real root. When the discriminant is less than zero, there are no real roots, but there are exactly two distinct imaginary roots. In this case, we have two real roots.

Finally, we use the quadratic function to find these exact roots. The quadratic function is:

$x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$

Plugging in our values of a, b, and c, we get:

$x = \frac{-19\pm \sqrt{19^2 - 4\cdot 4\cdot -5}}{2\cdot 4}$

This simplifies to:

$x = \frac{-19\pm \sqrt{441}}{8}$

which simplifies to

$x = \frac{-19\pm21}{8}$

which gives us two answers:

$x=\frac{1}{4}$ or x=-5

These values of x are the two distinct real roots of the given equation.

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Example Question #2 : Find Roots Of Quadratic Equation Using Discriminant

Use the formula b² - 4ac to find the discriminant of the following equation: 4x² + 12x + 10 = 0.

Then state how many roots it has, and whether they are real or imaginary. Finally, use the quadratic function to find the exact roots of the equation.

Possible Answers:

Discriminant: 304

Types of Roots: Two distinct real roots

Exact Roots: $\frac{3\pm \sqrt{19}}{2}$

Discriminant: 304

Types of Roots: Two distinct real roots

Exact Roots: $\frac{-3\pm \sqrt{19}}{2}$

Discriminant: -16

Types of Roots: No real roots; 2 distinct imaginary roots

Exact Roots: $x=\frac{-3\pm i}{2}$

Discriminant: -16

Types of Roots: No real roots; 2 distinct imaginary roots

Exact Roots: $x=\frac{3\pm i}{2}$

Discriminant: 16

Types of Roots: Two distinct real roots

Exact Roots: -1, -2

Correct answer:

Discriminant: -16

Types of Roots: No real roots; 2 distinct imaginary roots

Exact Roots: $x=\frac{-3\pm i}{2}$

Explanation:

In the above equation, a = 4, b = 12, and c = 10. Therefore:

b² - 4ac = (12)² - 4(4)(10) = 144 - 160 = -16.

Finally, we use the quadratic function to find these exact roots. The quadratic function is:

$x = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a}$

Plugging in our values of a, b, and c, we get:

$x = \frac{-12\pm \sqrt{144 - 4\cdot 4\cdot 10}}{2\cdot 4}$

This simplifies to:

$x = \frac{-12\pm \sqrt{-16}}{8}$

$x=\frac{-12\pm 4i}{8}$

$x=\frac{-3\pm i}{2}$

In other words, our two distinct imaginary roots are $\frac{-3+i}{2}$ and $\frac{-3-i}{2}$

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Precalculus : Graphing Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #91 : Graphing Functions

Example Question #11 : Distance

Example Question #31 : Calculus

Example Question #2 : Find The Distance Between Two Parallel Lines

Example Question #31 : Gre Subject Test: Math

Example Question #1 : Find Roots Of Quadratic Equation Using Discriminant

Example Question #2 : Find Roots Of Quadratic Equation Using Discriminant

Example Question #3 : Find Roots Of Quadratic Equation Using Discriminant

Example Question #1 : Find Roots Of Quadratic Equation Using Discriminant

Example Question #2 : Find Roots Of Quadratic Equation Using Discriminant

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