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Precalculus : Graphing Functions

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Example Questions

Example Question #4 : Find Roots Of Quadratic Equation Using Discriminant

Use the formula b² - 4ac to find the discriminant of the following equation: -3x² + 6x - 3 = 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: x = -1

Discriminant: 0

One real root: x = 1

Discriminant: 72

Two distinct real roots: $x=-1\pm \sqrt{2}$

Discriminant: 72

Two distinct real roots: $x=1\pm \sqrt{2}$

Discriminant: -72

Two distinct imaginary roots: $x=1\pm 2i$

Correct answer:

Discriminant: 0

One real root: x = 1

Explanation:

In the above equation, a = -3, b = 6, and c = -3. Therefore:

b² - 4ac = (6)² - 4(-3)(-3) = 36 - 36 = 0.

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, there is exactly one real root.

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

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

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

$x = \frac{-6\pm \sqrt{36 - 4\cdot -3\cdot -3}}{2\cdot -3}$

This simplifies to:

$x = \frac{-6\pm \sqrt{36 - 36}}{-6}$

which simplifies to

$x= \frac{-6\pm0}{-6}$

which gives us one answer: x = 1

This value of x is the one distinct real root of the given equation.

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

Use the formula b² - 4ac to find the discriminant of the following equation: x² + 5x + 4 = 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: 41

Two imaginary roots: $x=\frac{-5\pm 41i}{2}$

Discriminant: 0

One real root: $x=\frac{-5}{2}$

Discriminant: 9

Two real roots: x = -1 or x = -4

Discriminant: 9

Two real roots: x = 1 or x = 4

Discriminant: 41

Two imaginary roots: $x=\frac{5\pm 41i}{2}$

Correct answer:

Discriminant: 9

Two real roots: x = -1 or x = -4

Explanation:

In the above equation, a = 1, b = 5, and c = 4. Therefore:

b² - 4ac = (5)² - 4(1)(4) = 25 - 16 = 9.

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:

$x = \frac{-5\pm \sqrt{5^2 - 4\cdot 1\cdot 4}}{2\cdot 1}$

This simplifies to:

$x = \frac{-5\pm \sqrt{9}}{2}$

which simplifies to

$x=\frac{-5\pm3}{2}$

which gives us two answers:

x = -1 or x = -4

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

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Example Question #4 : Find Roots Of Quadratic Equation Using Discriminant

Use the formula b² - 4ac to find the discriminant of the following equation: -x² + 3x - 3 = 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: -21

Two imaginary roots: $x=\frac{-3\pm\sqrt{21}i}{2}$

Discriminant: -8

Two imaginary roots: $x=\frac{3\pm\sqrt{3}i}{2}$ .

Discriminant: 0

One real root: $x=\frac{3}{2}$

Discriminant: -8

Two imaginary roots: $x=\frac{-3\pm\sqrt{3}i}{2}$

Discriminant: -21

Two imaginary roots: $x=\frac{3\pm\sqrt{21}i}{2}$

Correct answer:

Discriminant: -8

Two imaginary roots: $x=\frac{3\pm\sqrt{3}i}{2}$ .

Explanation:

In the above equation, a = -1, b = 3, and c = -3. Therefore:

b² - 4ac = (3)² - 4(-1)(-3) = 9 - 12 = -3.

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:

$x = \frac{-3\pm \sqrt{3^2 - 4\cdot -1\cdot -3}}{2\cdot -1}$

This simplifies to:

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

Because $\sqrt{-3}=\sqrt{(-1)(3)}=\sqrt{3}i$ , this simplifies to $x=\frac{3\pm\sqrt{3}i}{2}$ . In other words, our two distinct imaginary roots are $x=\frac{3+\sqrt{3}i}{2}$ and $x=\frac{3-\sqrt{3}i}{2}$

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Example Question #5 : Find Roots Of Quadratic Equation Using Discriminant

Use the formula b² - 4ac to find the discriminant of the following equation: x² + 2x + 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: 36

Two real roots: x = 5 or x = -7

Discriminant: -36

Two imaginary roots: $-1\pm3i$

Discriminant: 0

One real root: x = -1

Discriminant: -36

Two imaginary roots: $1\pm3i$

Discriminant: 36

Two real roots: x = -5 or x = 7

Correct answer:

Discriminant: -36

Two imaginary roots: $-1\pm3i$

Explanation:

In the above equation, a = 1, b = 2, and c = 10. Therefore:

b² - 4ac = (2)² - 4(1)(10) = 4 - 40 = -36.

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:

$x = \frac{-2\pm \sqrt{2^2 - 4\cdot 1\cdot 10}}{2\cdot 1}$

This simplifies to:

$x = \frac{-2\pm \sqrt{-36}}{2}$

Because $\sqrt{-36}=\sqrt{(-1)(36)}=6i$ , this simplifies to $x=\frac{-2\pm6i}{2}$ . We can further simplify this to $-1\pm3i$ . In other words, our two distinct imaginary roots are -1+3i and -1-3i .

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Example Question #6 : Find Roots Of Quadratic Equation Using Discriminant

Use the formula b² - 4ac to find the discriminant of the following equation: x² + 8x + 16 = 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: x = 0

Discriminant: 0

One real root: x = -4

Discriminant: 128

Two distinct real roots: $x=-4\pm4\sqrt{2}$

Discriminant: 0

One real root: x = 4

Discriminant: 72

Two distinct real roots: $x=4\pm4\sqrt2$

Correct answer:

Discriminant: 0

One real root: x = -4

Explanation:

In the above equation, a = 1, b = 8, and c = 16. Therefore:

b² - 4ac = (8)² - 4(1)(16) = 64 - 64 = 0.

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

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

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

$x = \frac{-8\pm \sqrt{8^2 - 4\cdot 1\cdot 16}}{2\cdot 1}$

This simplifies to:

$x = \frac{-8\pm \sqrt{0}}{2}$

which simplifies to

x=-4

This value of x is the one distinct real root of the given equation.

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Example Question #1 : Amplitude, Period, Phase Shift Of A Trig Function

What is the amplitude of the following function?

$y = -24\sin(x-\pi)+14$

Possible Answers:

$\pi$

-24

-14

Correct answer:

Explanation:

When you think of a trigonometric function of the form y=Asin(Bx+C)+D, the amplitude is represented by A, or the coefficient in front of the sine function. While this number is -24, we always represent amplitude as a positive number, by taking the absolute value of it. Therefore, the amplitude of this function is 24.

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Example Question #101 : Graphing Functions

Select the answer choice that correctly matches each function to its period.

Possible Answers:

$\sin x \rightarrow \pi$

$\cos x \rightarrow \pi$

$\tan x \rightarrow \pi$

$\cot x \rightarrow \pi$

$\sec x \rightarrow \pi$

$\csc x \rightarrow \pi$

$\sin x \rightarrow 2\pi$

$\cos x \rightarrow 2\pi$

$\tan x \rightarrow \pi$

$\cot x \rightarrow \pi$

$\sec x \rightarrow 2\pi$

$\csc x \rightarrow 2\pi$

$\sin x \rightarrow 2\pi$

$\cos x \rightarrow 2\pi$

$\tan x \rightarrow 2\pi$

$\cot x \rightarrow 2\pi$

$\sec x \rightarrow \pi$

$\csc x \rightarrow \pi$

$\sin x \rightarrow \pi$

$\cos x \rightarrow \pi$

$\tan x \rightarrow 2\pi$

$\cot x \rightarrow 2\pi$

$\sec x \rightarrow 2\pi$

$\csc x \rightarrow 2\pi$

$\sin x \rightarrow 2\pi$

$\cos x \rightarrow 2\pi$

$\tan x \rightarrow 2\pi$

$\cot x \rightarrow 2\pi$

$\sec x \rightarrow 2\pi$

$\csc x \rightarrow 2\pi$

Correct answer:

$\sin x \rightarrow 2\pi$

$\cos x \rightarrow 2\pi$

$\tan x \rightarrow \pi$

$\cot x \rightarrow \pi$

$\sec x \rightarrow 2\pi$

$\csc x \rightarrow 2\pi$

Explanation:

The following matches the correct period with its corresponding trig function:

$\sin x \rightarrow 2\pi$

$\cos x \rightarrow 2\pi$

$\tan x \rightarrow \pi$

$\cot x \rightarrow \pi$

$\sec x \rightarrow 2\pi$

$\csc x \rightarrow 2\pi$

In other words, sin x, cos x, sec x, and csc x all repeat themselves every $2\pi$ units. However, tan x and cot x repeat themselves more frequently, every $\pi$ units.

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Example Question #2 : Amplitude, Period, Phase Shift Of A Trig Function

What is the period of this sine graph?

Trig period 1

Possible Answers:

$3\pi$

$\pi^2$

$\frac{3}{2}\pi$

$\frac{2}{3}\pi$

Correct answer:

$\frac{2}{3}\pi$

Explanation:

The graph has 3 waves between 0 and $2\pi$ , meaning that the length of each of the waves is $2\pi$ divided by 3, or $\frac{2}{3}\pi$ .

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Example Question #3 : Amplitude, Period, Phase Shift Of A Trig Function

Write the equation for a cosine graph with a maximum at $\left( \frac{ \pi }{3} , 3 \right)$ and a minimum at $\left(\frac{ 4 \pi }{3} , -3 \right)$ .

Possible Answers:

$y = 2 \cos (x + \frac{ \pi }{3} )+ 1$

$y = 3 \cos (x - \frac{ \pi }{3} )$

$y = -3 \cos (x - \frac{ \pi } {3 })$

$y = 3 \cos (\frac{ 1}{2} ( x - \frac{ \pi }{3} ))$

$y = 2 \cos (\frac{ 1}{2} ( x + \frac{ \pi }{3} )) + 1$

Correct answer:

$y = 3 \cos (x - \frac{ \pi }{3} )$

Explanation:

In order to write this equation, it is helpful to sketch a graph:

Trig graph 2

The dotted line is at $x = \frac{ \pi }{3}$ , where the maximum occurs and therefore where the graph starts. This means that the graph is shifted to the right $\frac{ \pi }{3}$ .

The distance from the maximum to the minimum is half the entire wavelength. Here it is $\frac{ 4 \pi }{3} - \frac{ \pi }{3} = \frac{ 3 \pi }{3} = \pi$ .

Since half the wavelength is $\pi$ , that means the full wavelength is $2 \pi$ so the frequency is just 1.

The amplitude is 3 because the graph goes symmetrically from -3 to 3.

The equation will be in the form $y = A \cdot \cos (f (x - h)) + k$ where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This equation is

$y = 3 \cos (x - \frac{ \pi } {3 })$ .

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Example Question #102 : Graphing Functions

Find the phase shift of F(t)=3sin(2t+4) .

Possible Answers:

-4

-2

Correct answer:

-2

Explanation:

In the formula,

F(t)=Asin(Bt-C)+D .

$\frac{C}{B}$ represents the phase shift.

Plugging in what we know gives us:

$\frac{C}{B}=\frac{-4}{2}$ .

Simplified, the phase is then .

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Precalculus : Graphing Functions

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #4 : Find Roots Of Quadratic Equation Using Discriminant

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

Example Question #4 : Find Roots Of Quadratic Equation Using Discriminant

Example Question #5 : Find Roots Of Quadratic Equation Using Discriminant

Example Question #6 : Find Roots Of Quadratic Equation Using Discriminant

Example Question #1 : Amplitude, Period, Phase Shift Of A Trig Function

Example Question #101 : Graphing Functions

Example Question #2 : Amplitude, Period, Phase Shift Of A Trig Function

Example Question #3 : Amplitude, Period, Phase Shift Of A Trig Function

Example Question #102 : Graphing Functions

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