SAT Math - Graphing Linear Functions

Determine where the graphs of the following equations will intersect.

Explanation

We can solve the system of equations using the substitution method.

Solve for in the second equation.

Substitute this value of into the first equation.

Now we can solve for .

$y=\frac{14}{-7}=-2$

Solve for using the first equation with this new value of .

$x=\frac{3}{3}=1$

The solution is the ordered pair .

Inequality

Which of the following inequalities is graphed above?

$y \leq - \frac{3}{8} x + 3$

$y \geq - \frac{3}{8} x + 3$

$y \leq - \frac{3}{8} x +8$

$y \geq - \frac{8}{3} x - 3$

$y \geq - \frac{8}{3} x - 8$

Explanation

First, we determine the equation of the boundary line. This line includes points and , so the slope can be calculated as follows:

$m = \frac{y_{2}- y_{1}}{x_{2}- x_{1}} = \frac{3-0}{0- 8} = - \frac{3}{8}$

Since we also know the -intercept is , we can substitute $m = - \frac{3}{8}, b = 3$ in the slope-intercept form to obtain the equation of the boundary line:

$y = - \frac{3}{8} x + 3$

The boundary is included, as is indicated by the line being solid, so the equality symbol is replaced by either $\leq$ or $\geq$ . To find out which one, we can test a point in the solution set - for ease, we will choose :

_____ $- \frac{3}{8} x + 3$

_____ $- \frac{3}{8}\cdot 0 + 3$

_____

0 is less than 3 so the correct symbol is $\leq$ .

The inequality is $y \leq - \frac{3}{8} x + 3$ .

Screenshot_from_2014-03-27_16_11_40

Which equation best matches the graph of the line shown above?

$y+2=\frac{1}{4}(x-4)$

$y-2=\frac{1}{4}(x-4)$

Explanation

To find an equation of a line, we will always need to know the slope of that line -- and to find the slope, we need at least two points. It looks like we have (0, -3) and (12,0), which we'll call point 1 and point 2, respectively.

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{0-(-3)}{12-0}=\frac{3}{12}=\frac{1}{4}$

Now we need to plug in a point on the line into an equation for a line. We can use either slope-intercept form or point-slope form, but since the answer choices are in point-slope form, let's use that.

$y=\frac{1}{4}(x-12)$

Unfortunately, that's not one of the answer choices. That's because we didn't pick the same point to substitute into our equation as the answer choices did. But we can see if any of the answer choices are equivalent to what we found. Our equation is equal to:

$y=\frac{1}{4}x-3$

which is the slope-intercept form of the line. We have to put all the other answer choices into slope-intercept to see if they match. The only one that works is this one:

$y+2=\frac{1}{4}(x-4)$

$y=\frac{1}{4}(x-4)-2$

$y=\frac{1}{4}x-1-2$

$y=\frac{1}{4}x-3$

Axes

Refer to the above diagram. If the red line passes through the point $\left ( N, 4\right )$ , what is the value of ?

$N = -4\frac{2}{3}$

$N= -1\frac{1}{3}$

$N = -7\frac{1}{3}$

$N = -3\frac{1}{3}$

Explanation

One way to answer this is to first find the equation of the line.

The slope of a line. given two points $(x_{1}, y_{1}), (x_{2}, y_{2})$ can be calculated using the slope formula

$m = \frac{y_{2}-y_{1}}{x_{2}-x _{1}}$

Set $x_{1}=-6, y_{1}=x_{2}= 0, y_{2}=18$ :

$m = \frac{18-0}{0-(-6)} = \frac{18}{6} = 3$

The line has slope 3 and -intercept , so we can substitute in the slope-intercept form:

Now substitute 4 for and for and solve for :

$N = -\frac{14}{3}= -4\frac{2}{3}$

An individual's maximum heart rate can be found by subtracting his or her age from . Which graph correctly expresses this relationship between years of age and maximum heart rate?

Screen_shot_2015-02-14_at_6.24.06_pm

Screen_shot_2015-02-14_at_6.24.18_pm

Screen_shot_2015-02-14_at_6.24.40_pm

Screen_shot_2015-02-14_at_6.31.38_pm

Screen_shot_2015-02-14_at_6.31.44_pm

Explanation

In form, where y = maximum heart rate and x = age, we can express the relationship as:

We are looking for a graph with a slope of -1 and a y-intercept of 220.

The slope is -1 because as you grow one year older, your maximum heart rate decreases by 1.

Select the equation of the line perpendicular to the graph of .

$y=\frac{4}{3}x-7$

$y=\frac{3}{4}x-8$

$y=-\frac{3}{4}x+8$

$y=-\frac{4}{3}x-7$

None of these.

Explanation

Lines are perpendicular when their slopes are the negative recicprocals of each other such as $2\hspace{1mm}and\hspace{1mm}-\frac{1}{2}$ . To find the slope of our equation we must change it to slope y-intercept form.

Subtract the x variable from both sides:

Divide by 4 to isolate y:

$y=\frac{-3}{4}x+2$

The negative reciprocal of the above slope: $\frac{-3}{4} \rightarrow \frac{4}{3}$ . The only equation with this slope is $y=\frac{4}{3}x-7$ .

Axes_1

Refer to the line in the above diagram. It we were to continue to draw it so that it intersects the -axis, where would its -intercept be?

$\left ( 0, - 1\frac{1}{3}\right )$

$\left ( 0, - 1\frac{2}{3}\right )$

$\left ( 0, - 1\right )$

$\left ( 0, - 1\frac{3}{5}\right )$

$\left ( 0, - 1\frac{2}{5}\right )$

Explanation

First, we need to find the slope of the line.

In order to move from the lower left point to the upper right point, it is necessary to move up five units and right three units. This is a rise of 5 and a run of 3. makes the slope of the line shown $\frac{5}{3}$ .

We can use this to find the -intercept using the slope formula as follows:

The lower left point has coordinates . Therefore, we can set up and solve for in this slope formula, setting $x_{1} = y_{1} = 2, x_{2} = 0, y_{2} = b, m = \frac{5}{3}$ :

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = m$

$\frac{b-2}{0-2} = \frac{5}{3}$

$\frac{b-2}{ -2} = \frac{5}{3}$

$\frac{b-2}{ -2} \cdot (-2)= \frac{5}{3}\cdot (-2)$

$b-2 = -\frac{10}{3}$

$b-2 + 2 = -\frac{10}{3} + 2$

$b = -\frac{4}{3} =-1 \frac{1}{3}$

Line

Refer to the above red line. A line is drawn perpendicular to that line, and with the same -intercept. Give the equation of that line in slope-intercept form.

$y = - \frac{1}{2}x + 8$

$y = \frac{1}{2}x + 8$

$y = - \frac{1}{2}x + 4$

$y = \frac{1}{2}x + 4$

$y = \frac{1}{2}x - 4$

Explanation

First, we need to find the slope of the above line.

The slope of a line. given two points $(x_{1}, y_{1}), (x_{2}, y_{2})$ can be calculated using the slope formula

$m = \frac{y_{2}-y_{1}}{x_{2}-x _{1}}$

Set $x_{1}=-4, y_{1}=x_{2}= 0, y_{2}=8$ :

$m = \frac{8-0}{0-(-4)} = \frac{8}{4} = 2$

The slope of a line perpendicular to it has as its slope the opposite of the reciprocal of 2, which would be $m = -\frac{1}{2}$ . Since we want this line to have the same -intercept as the first line, which is the point , we can substitute $m = -\frac{1}{2}$ and in the slope-intercept form:

$y = - \frac{1}{2}x + 8$

Line includes the points and . Line includes the points and . Which of the following statements is true of these lines?

The lines are parallel.

The lines are perpendicular.

The lines are identical.

The lines are distinct but neither parallel nor perpendicular.

Insufficient information is given to answer this question.

Explanation

We calculate the slopes of the lines using the slope formula.

The slope of line is

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{4-4}{7 - (-3)} = \frac{0}{10} =0$

The slope of line is

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{-4- \left ( -4\right )}{8 - 5} = \frac{0}{3} =0$

The lines have the same slope, making them either parallel or identical.

Since the slope of each line is 0, both lines are horizontal, and the equation of each takes the form , where is the -coordinate of each point on the line. Therefore, line and line have equations and .This makes them parallel lines.

Circle

Note: Figure NOT drawn to scale.

Refer to the above figure. The circle has its center at the origin; the line is tangent to the circle at the point indicated. What is the equation of the line in slope-intercept form?

$y = - \frac{3}{4} x+ \frac{25}{2}$

Insufficient information is given to determine the equation of the line.

$y = \frac{3}{4} x+ \frac{7}{2}$

$y = - \frac{4}{3} x+ 16$

$y = \frac{4}{3} x$

Explanation

A line tangent to a circle at a given point is perpendicular to the radius from the center to that point. That radius, which has endpoints , has slope

$m = \frac{8-0}{6-0} = \frac{8}{6} = \frac{4}{3}$ .

The line, being perpendicular to this radius, will have slope equal to the opposite of the reciprocal of that of the radius. This slope will be $-\frac{3}{4}$ . Since it includes point , we can use the point-slope form of the line to find its equation: