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Precalculus : Linear Functions

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Example Questions

Example Question #1 : Understanding Complementary And Suplmentary Angles

Solve for .

(Figure not drawn to scale).

Possible Answers:

12^o

10^o

8^o

14^o

Correct answer:

14^o

Explanation:

The angles are supplementary, therefore, the sum of the angles must equal 180^o .

$\small (4n+22^o)+(8n-10^o)=180^o$

$\small 4n+22^o+8n-10^o=180^o$

$\small 12n+12^o=180^o$

$\small 12n=168^o$

$\small n=14^o$

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Example Question #2 : Understanding Complementary And Suplmentary Angles

Are $129^{\circ}$ and $51^{\circ}$ supplementary angles?

Possible Answers:

Yes

Not enough information

Correct answer:

Yes

Explanation:

Since supplementary angles must add up to $180^{\circ}$ , the given angles are indeed supplementary.

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Example Question #1 : Finding Angles

Solve for and .

(Figure not drawn to scale).

Possible Answers:

$\small x=15^o;\ y=37.5^o$

$\small x=15^o;\ y=52.5^o$

$\small x=7.5^o;\ y=18.75^o$

$\small x=15^o;\ y=7.5^o$

Correct answer:

$\small x=15^o;\ y=37.5^o$

Explanation:

The angles containing the variable all reside along one line, therefore, their sum must be 180^o .

$\small 5x+4x+3x=180^o$

$\small 12x=180^o$

$\small x=15^o$

Because and are opposite angles, they must be equal.

$\small 2y=5x$

$\small x=15^o$

$\small 2y=5(15^o)=75^o$

$\small y=\frac{75^o}{2}=37.5^o$

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Example Question #1 : Graph A Linear Function

Which of the following could be the function modeled by this graph?

Linearfxn

Possible Answers:

y=7x-5

y=5x-7

y=5x+7

$y=\frac{1}{5}x-7$

Correct answer:

y=5x-7

Explanation:

Which of the following could be the function modeled by this graph?

Linearfxn

We can begin here by trying to identify a couple points on the graph

We can see that it crosses the y-axis at (0,-7)

Therefore, not only do we have a point, we have the y-intercept. This tells us that the equation of the line needs to have a in it somewhere. Eliminate any option that do not have this feature.

Next, find the slope by counting up and over from the y-intercept to the next clear point.

It seems like the line goes up 5 and right 1 to the point (1,-2)

This means we have a slope of 5, which means our equation must look like this:

y=5x-7

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Example Question #1 : Graph A Linear Function

Find the slope of the linear function

y-3=5(x+2)

Possible Answers:

m=2

m=-3

m=3

m=5

Correct answer:

m=5

Explanation:

For the linear function in point-slope form

$y-y_{1}=m\left ( x-x_{1}\right )$

The slope is equal to

For this problem

y-3=5(x+2)

we get

m=5

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Example Question #2 : Graph A Linear Function

Find the slope of the linear function

y-7=10(x+1)

Possible Answers:

m=-7

m=1

m=10

m=7

Correct answer:

m=10

Explanation:

For the linear function in point-slope form

$y-y_{1}=m\left ( x-x_{1}\right )$

The slope is equal to

For this problem

y-7=10(x+1)

we get

m=10

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Example Question #3 : Linear Functions

What is the y-intercept of the line below?

y = 3x + 4

Possible Answers:

(3,0)

(0, 4)

(4,0)

(0, -3)

(0, 3)

Correct answer:

(0, 4)

Explanation:

By definition, the y-intercept is the point on the line that crosses the y-axis. This can be found by substituting x = 0 into the equation. When we do this with our equation,

$y = 3x + 4 = 3\cdot0 + 4 = 0 + 4 = 4$ .

Alternatively, you can remember y = mx + b form, a general form for a line in which is the slope and is the y-intercept.

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Example Question #4 : Linear Functions

What is the slope of the line below?

$y = \frac{2}{3}x + 6$

Possible Answers:

$\frac{1}{6}$

$-\frac{2}{3}$

$\frac{2}{3}$

Correct answer:

$\frac{2}{3}$

Explanation:

Recall slope-intercept form, or y = mx + b . In this form, is the slope and is the y-intercept. Given our equation above, the slope must be the coefficient of the x, which is $\frac{2}{3}$ .

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Example Question #5 : Linear Functions

What is the x-intercept of the equation below?

y = 3x - 6

Possible Answers:

(0, -6)

$(\frac{1}{2}, 0)$

(-2, 0)

(0, 2)

(2,0)

Correct answer:

(2,0)

Explanation:

The x-intercept of an equation is the point at which the line crosses the x-axis. Thus, we can find the x-intercept by plugging in y=0 . When we do this with our equation:

$y = 3x - 6 \rightarrow 0 = 3x - 6\rightarrow6 = 3x \rightarrow2 = x$

Thus, our x-intercept is the point (2,0) .

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Example Question #1 : Determine The Equation Of A Linear Function

What is the equation of the line that passes through the points (0,2) and (4,7) ?

Express your answer in y=mx+b form.

Possible Answers:

y =2

None of the other answers.

$y = \frac{5}{4}x+2$

y = 4x+2

$y = \frac{1}{2}x + 5$

Correct answer:

$y = \frac{5}{4}x+2$

Explanation:

First, we need to compute , the slope. We can do this with the slope formula

$m = \frac{y_2-y_1}{x_2-x_1}$ , sometimes called "rise over run"

$m = \frac{7-2}{4-0} = \frac{5}{4}$

So we now have

$y = \frac{5}{4}x +b$

Now in order to solve for we substitute one of our points into the equation we found. It doesn't matter which point we use, so we'll use (0,2) .

We then have:

$2 = \frac{5}{4}0 +b$

Which becomes 2 = b .

Hence we take our found value for and plug it back into $y= \frac{5}{4}x+b$ to get

$y =\frac{5}{4}x+2$ .

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Precalculus : Linear Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Understanding Complementary And Suplmentary Angles

Example Question #2 : Understanding Complementary And Suplmentary Angles

Example Question #1 : Finding Angles

Example Question #1 : Graph A Linear Function

Example Question #1 : Graph A Linear Function

Example Question #2 : Graph A Linear Function

Example Question #3 : Linear Functions

Example Question #4 : Linear Functions

Example Question #5 : Linear Functions

Example Question #1 : Determine The Equation Of A Linear Function

All Precalculus Resources

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