Chapter 1 Homework Pre-Cal (1.1 - 1.5)

1.1 Match the story with the correct graph. As the storm got worse, the rain got harder and harder.

To match the story with the correct​ graph, first determine what is happening with amount of rain as time goes by. Notice that ​"the rain got harder and ​harder" so the amount of rain is constantly increasing. Remember that the amount of rain is increasing when the graph rises from left to right and the amount of rain stays constant when the graph is the horizontal line. ​Thus, a is the correct graph.

1.5 Find the average rate of change of the function f(x)=x2+5x from x1=3 to x2=8.

Recall that the average rate of change of a function​ f(x) from x1 to x2 is fx2−fx1x2−x1. We are given x1=3 and x2=8. Evaluate the function ​f(x)=x2+5x at the following values. f(3)=32+5(3)=24 f(8)=82+5(8)=104 Substitute the values into the formula and calculate. Simplify your result. f(8)−f(3)8−3 = 104−248−3 Substitute. f(8)−f(3)8−3 = 805 Subtract. f(8)−f(3)8−3 = 16 Divide. The average rate of change is 16.

1.2 Determine whether the following equation defines y as a function of x. x^2+y^2=100 Does the equation x^2+y^2=100 define y as a function of​ x? Yes No

No

1.2 Determine whether the relation is a function. Give the domain and the range for the relation. ​{(3​,2​),(4​,7​),(8​,8​)}

A relation is any set of ordered pairs. The set of all first components of the ordered pair is called the domain of the​ relation, and the set of all second components is called the range of the relation. The domain is the set of all first components.​ Thus, the domain is ​{3​,4​,8​}. The range is the set of all second components.​ Thus, the range is ​{2​,7​,8​}. A function is a relation in which each member of the domain corresponds to exactly one member of the range. No two ordered pairs of a function can have the same first component and different second components. Look at the first and second components of the ordered pairs in the given relation. Notice that no two ordered pairs have the same first component and different second components. ​Therefore, the relation is a function. The domain is ​{3​,4​,8​} and the range is ​{2​,7​,8​}.

1.1 Plot the point (−7/2,5/2) in a rectangular coordinate system.

A set of perpendicular number lines divides the plane into four quadrants. The intersection of the axes is the​ origin, and the quadrants are numbered as shown. Each point in a rectangular coordinate system corresponds to an ordered pair of real​ numbers, (x,y). The first number in each ordered​ pair, called the​ x-coordinate, denotes the distance and direction from the origin along the​ x-axis. A positive​ x-coordinate is shown to the right of the​ origin, while a negative​ x-coordinate is shown to the left of the origin. The second​ number, called the​ y-coordinate, denotes vertical distance and direction along a line parallel to the​ y-axis or along the​ y-axis itself. A positive​ y-coordinate is shown above the​ origin, while a negative​ y-coordinate is shown below the origin. Start by identifying the x and​ y-coordinates of the point −7/2,5/2. ​x-coordinate=−7/2 ​y-coordinate=5/2 Find the decimal equivalents of the coordinates to help plot the point. −7/2,5/2=(−3.5,2.5) To plot (−3.5,2.5)​, start at the origin. Go 3.5 units to the left because the​ x-coordinate is negative. Notice that the tick size is 0.5. Go 2.5 units up parallel to the​ y-axis because the​ y-coordinate is positive. The point −7/2,5/2 is shown on the coordinate system to the right.

1.5 Write the​ slope-intercept equation of the function f whose graph satisfies the given conditions. The graph of f passes through (4,6) and is perpendicular to the line whose equation is x=2.

A vertical line is given by an equation of the form x=a​, where a is the​ x-intercept. The line x=2 is of the form x=a, so it is a vertical line. A line perpendicular to x=2 will intersect x=2 at a right angle. The only kind of line that will intersect a vertical line at a right angle is a horizontal​ line, y=b. All points on a horizontal line have the same​ y-coordinate. The b in the equation of the horizontal line is the​ y-coordinate of all points on the horizontal line. The​ y-coordinate of the point (4,6) is 6. The equation of the function whose graph is perpendicular to the line x=2 and contains the point (4,6) is

1.5 Suppose that a ball is rolling down a ramp. The distance traveled by the ball is given by the function s(t)=10t2​, where t is the​ time, in​ seconds, after the ball is​ released, and​ s(t) is measured in feet. Find the​ ball's average velocity in each of the following time intervals.

Average Velocity of an Object Suppose that a function expresses an​ object's position,​ s(t), in terms of​ time, t. The average velocity of the object from t1 to t2 is ΔsΔt=st2−st1t2−t1. a. t1=3 to t2=4 ΔsΔt = s(4)−s(3)4 sec−3 sec Calculate​ s(t) for both values of t. = 160 ft−90 ft1 sec Simplify. = 70 ft/sec b. t1=3 to t2=3.5 ΔsΔt = s(3.5)−s(3)3.5 sec−3 sec Calculate​ s(t) for both values of t. = 122.5 ft−90 ft0.5 sec Simplify. = 65 ft/sec c. t1=3 to t2=3.01 ΔsΔt = s(3.01)−s(3)3.01 sec−3 sec Calculate​ s(t) for both values of t. Find the exact values. = 90.601 ft−90 ft0.01 sec Simplify. Find the exact value. = 60.1 ft/sec d. t1=3 to t2=3.001 ΔsΔt = s(3.001)−s(3)3.001 sec−3 sec Calculate​ s(t) for both values of t. Find the exact values. = 90.06001 ft−90 ft0.001 sec Simplify. Find the exact value. = 60.01 ft/sec

1.4 Find the value of y if the line through the two given points is to have the indicated slope. (3,y) and (2,4), m=−5

For this​ problem, use the definition of slope to find the value of y. Recall that the slope of the line through the distinct points x1,y1 and x2,y2 is defined as the following. Change in yChange in x ​= RiseRun ​= y2−y1x2−x1 Note that x2−x1 cannot equal 0. Let x1,y1=(3,y) and x2,y2=(2,4). With the slope being −5, substitute these values into the slope equation. Slope ​= y2−y1x2−x1 −5 ​= (4)−(y)(2)−(3) Simplify the denominator of the expression. −5 ​= 4−y−1 Multiply both sides of the equation by the​ denominator, −1. 5 ​= 4−y Subtract 4 from both sides of the equation and simplify. 1 ​= −y Multiply both sides by −1 to find the value of y. −1 ​= y

1.3 Use the graph to determine a. open intervals on which the function is​ increasing, if any. b. open intervals on which the function is​ decreasing, if any. c. open intervals on which the function is​ constant, if any. A coordinate system has a horizontal x-axis labeled from negative 5 to 5 in increments of 1 and a vertical y-axis labeled from negative 5 to 5 in increments 1. A parabola that opens downward has vertex (3, 2) and passes through the approximate points (2, 1.7) and (4, 1.7).

Increasing and Decreasing Functions: 1. A function is increasing on an open​ interval, I​, if for any x1 and x2 in the​ interval, where x1<x2 then fx1<fx2. 2. A function is decreasing on an open​ interval, I​, if for any x1 and x2 in the​ interval, where x1<x2 then fx1>fx2. Constant Functions: 3. A function is constant on an open​ interval, I​, if for any x1 and x2 in the​ interval, where x1<x2 then fx1=fx2. a. Determine the open intervals on which the function is​ increasing, if any. There is one open interval on which the function is increasing. The open interval starts at x=−∞. The open interval ends at x=3. The function increases on the open interval (−∞,3). That​ is, for any x1 and x2 in the open​ interval, if x1<x2​, then fx1<fx2. The function has a lower value at the smaller value of the​ x-coordinate. b. Determine the open intervals on which the function is​ decreasing, if any. There is one open interval on which the function is decreasing. The open interval starts at x=3. The open interval ends at x=∞. The function decreases on the open interval (3,∞). That​ is, for any x1 and x2 in the open​ interval, if x1<x2, then fx1>fx2. The function has higher value at the smaller value of​ x-coordinate. c. Determine the intervals on which the function is​ constant, if any. There are no intervals on which the function is constant. The graph smoothly​ increases, then smoothly​ decreases, without a​ "flat" section in between. That​ is, it is not possible to identify an open interval containing values x1 and x2 where x1<x2, and fx1=fx2. The function is increasing on the open interval (−∞,3) and decreasing on the open interval (3,∞).

1.2 Determine whether the following equation defines y as a function of x. x=y^2 Does the equation x=y^2 define y as a function of​ x? Yes No

No Note: To determine whether the equation x=y^2 defines y as a function of​ x, solve the equation for y in terms of x. If two or more values of y can be obtained for a given​ x, the equation is not a function. ​ If, for each value of​ x, there is one and only one value of​ y, the equation is a function of x.

1.4 Use the given conditions to write an equation for the line in​ point-slope form and in​ slope-intercept form. Slope​ = 3/10​, passing through the origin

Point-Slope Form of the Equation of a Line The ​point-slope equation of a nonvertical line with slope m that passes through the point x1, y1 is y−y1=mx−x1. ​a) In order to use the​ point-slope equation, we need the slope of the line and the coordinates of one point on the line. We know that the slope is 310​, and that the line passes through the​ origin, which has the coordinates of​ (0, 0). Now substitute 310 for​ m, 0 for x1​, and 0 for y1 into the​ point-slope equation of a line. y−y1=mx−x1 This is the​ point-slope form of the equation. y−0=310(x−0) Substitute. The result is y−0=310(x−0)​, which is the​ point-slope form of the equation of the line described above. ​Slope-Intercept Form of the Equation of a Line The ​slope-intercept equation of a nonvertical line with slope m and​ y-intercept b is y=mx+b. ​b) Notice that in the​ slope-intercept form, the y is isolated on the left side. Solve the​ point-slope equation y−0=310(x−0) for y by eliminating the 0 on the left side and using the distributive property on the right side. y−0=310(x−0) ​Point-slope form of the equation of the line. y=310x Simplify the left side of the equation. Use the distributive property on the right side. The result is y=310x​, which is the​ slope-intercept form of the equation of the given line.

1.2 Determine whether the equation defines y as a function of x. x^3+y^7=11

Solve the equation for y in terms of x. If two or more values of y can be obtained for a given​ x, the equation is not a function. Solve for y by subtracting x3 from both sides of the equation. x^3+y^7=11 This is the given equation. x^3+y^7−x^3 = 11−x^3 Subtract x^3. y^7= 11−x^3 Combine like terms. y=7 sqrt 11−x3 Extract the root from both sides of the equation. There is one and only one value of y for each value of x. For this​ reason, the given equation defines y as a function of x.

1.5 Write the​ slope-intercept equation of the function f whose graph satisfies the given conditions. The graph of f passes through (1,2) and is perpendicular to the line whose equation is x=−4. The equation of the function is _____

f(x)=2

1.1 Find seven ordered pairs to the equation y=x^2−6. Then determine its graph.

The given equation involves two​ variables, x and y.​ However, because variable x is​ squared, it is not a linear equation in two variables. Although the graph is not a​ line, it is still a picture of all the​ ordered-pair solutions of y=x2−6. ​Thus, we can use the​ point-plotting method to obtain the graph. Find several ordered pairs that are solutions of the equation. Start with x=−3. Substitute x=−3 into the equation y=x2−6 and solve for y. Then continue to do the same with each integer value until you reach x=3. x y=x2−6 −3 y=(−3)2−6=3 ​Next, substitute x=−2 into the equation y=x2−6 and solve for y. x y=x2−6 −2 y=(−2)2−6=−2 The remaining values are computed and shown below. −1 y=(−1)2−6=−5 0 y=(0)2−6=−6 1 y=(1)2−6=−5 2 y=(2)2−6=−2 3 y=(3)2−6=3 The complete table of values is shown below. To graph the​ equation, we will plot each set of​ (x,y) values. We begin by plotting (−3,3) on the rectangular coordinate system to the right. To do​ so, move 3 units to the left of the origin and then 3 units up. Next we plot (−2,−2). Move 2 units to the left of the origin and 2 units down. The remaining points have been plotted using the same method as the previous two steps. Now connect the points with a smooth curve to graph y=x2−6.

1.5 Write the​ slope-intercept equation of the function f whose graph satisfies the given conditions. The graph of f passes through (8,5) and is perpendicular to the line whose equation is x=1. The equation of the function is ___

f(x)=5 Note: ​Remember, a vertical line is given by an equation of the form x=a. Think carefully about what type of line would be perpendicular to a vertical​ line, and then use the given information to find the equation of one such line.

1.4 Write the​ point-slope form of the line satisfying the given conditions. Then use the​ point-slope form of the equation to write the​ slope-intercept form of the equation. Passing through ​(2​,8​) and ​(4​,16​)

The​ point-slope equation of a nonvertical line with slope m that passes through the point x1, y1 is y−y1=mx−x1. We start by using the two points ​(2​,8​) and (4, 16) to find the slope of the line. m=Change in yChange in x =16−84−2 =82 Subtract in the numerator and denominator. =4 Simplify to find the slope. Now determine the​ point-slope equation. Let x1, y1=(2,8). y−y1 = mx−x1 This is the​ point-slope form of the equation y−8 = 4(x−2) Substitute the given values. We now have the​ point-slope form of the equation of the given​ line, y−8=4(x−2). Note that we could have also used the ordered pair (4, 16) to find the​ point-slope form of the equation. In that​ case, the equation would be y−16=4(x−4). Solve the​ point-slope form of the equation for y and write an equivalent equation in​ slope-intercept form (y=mx+b). y−8 = 4(x−2) This is the​ point-slope equation. y−8 = 4x−8 Use the distributive property on the right side of the equation. y = 4x Add 8 to both sides. ​Therefore, the​ slope-intercept form of the​ line's equation is y=4x.

1.2 Determine whether the following equation defines y as a function of x. x^2+y=9

To determine whether the equation x2+y=9 defines y as a function of​ x, solve the equation for y in terms of x. If two or more values of y can be obtained for a given​ x, the equation is not a function. Begin by solving the equation for y. x2+y = 9 This is the given equation. y = 9−x2 Solve for y by subtracting x2 from both sides. From this last​ equation, we can see that for each value of​ x, there is one and only one value of y. ​ Thus, the equation defines y as a function of x.

1.2 Determine whether the following equation defines y as a function of x. x−7=y^2

To determine whether the equation x−7=y^2 defines y as a function of​ x, solve the equation for y in terms of x. If two or more values of y can be obtained for a given​ x, the equation is not a function. Begin by solving the equation for y using the square root method. x−7=y^2 This is the given equation. ±sqrt x−7=y Apply the square root property. The ± in the​ equation, y=±sqrt x−7​, shows that for any value of x greater than​ 7, there are two values of y. For this​ reason, the equation does not define y as a function of x.

1.4 Use the given conditions to write an equation for the line in​ point-slope form and​ slope-intercept form. Slope=−8​, passing through (−2,0)

To find the equation of a​ line, given the slope and a point on the​ line, use the​ point-slope form of an equation of a line. The​ point-slope form of the equation of a nonvertical line with slope m that passes through the point x1,y1 is y−y1=mx−x1. An equation of the line with slope −8 and containing the point (−2,0) can be found by using the​ point-slope form with m=−8, x1=−2, and y1=0. y−y1 ​= m(x−x1) y−(0) ​= −8(x−(−2)) Simplify to find the​ point-slope form of the equation. y−(0) ​= −8(x−(−2)) y−0 ​= −8(x+2) The​ slope-intercept form of an equation of a line is y=mx+b. To express the equation in the​ slope-intercept form, perform the distribution on the right side of the equation. Perform the distribution on the right side of the equation. y ​= −8(x+2) y ​= −8x−16

1.4 Graph the following equation in a rectangular coordinate system. f(x)=6

To graph the line whose equation is f(x)=6​, first identify the slope and the​ y-intercept. When the equation of a line is written in​ slope-intercept form, it is easy to find the slope and the​ y-intercept of the line. The​ slope-intercept form of an equation of a line with slope m and​ y-intercept b is y=mx+b. Notice that there is no​ x-term in the given equation.​ Therefore, another way to write f(x)=6 is f(x)=0x+6. The coefficient of x is the slope. The constant is the​ y-intercept. The slope of the line with the equation f(x)=0x+6 is m=0. The​ y-intercept of the line with the equation f(x)=0x+6 is b=6. To graph the​ line, start by graphing the​ y-intercept. Since the slope is​ 0, the line is horizontal. We complete the graph by drawing the horizontal line that contains this point.

1.1 Write the English sentence as an equation in two variables. Then graph the equation. The​ y-value is six more than twice the​ x-value.

We begin by translating the English sentence to an equation in two variables. The first phrase in the​ sentence, "The​ y-value," simply means that one of our variables is y. The word​ "is" corresponds to the = symbol. ​ Thus, the first part of the sentence translates as shown below. The​ y-value is six more than twice the​ x-value. y= The phrase ​"twice the​ x-value" is represented by the expression 2x. The phrase ​"six more than twice the​ x-value" indicates that we want to add 6 to 2x. ​Therefore, the equation that corresponds to the sentence is y=2x+6. The​ y-value is six more than twice the​ x-value. y= 2x+6 Now we want to graph the equation. To graph y=2x+6​, we will use the​ point-plotting method. The first step is to find ordered pairs that satisfy the equation. Then we plot these ordered pairs as points in the rectangular coordinate system and connect the points. For each value of​ x, we find the corresponding value for y. x y=2x+6 −3 y=2(−3)+6=0 −2 y=2(−2)+6=2 −1 y=2(−1)+6=4 0 y=2(0)+6=6 1 y=2(1)+6=8 2 y=2(2)+6=10 3 y=2(3)+6=12 We plot each of the ordered pairs on the coordinate grid to the right. A coordinate system has a horizontal x-axis labeled from negative 5 to 5 in increments of 1 and a vertical y-axis labeled from negative 20 to 20 in increments of 1. The following points are plotted (negative 3, 0), (negative 2, 2), (negative 1, 4), (0, 6), (1, 8), (2, 10), and (3, 12). We now connect the points with a line to obtain the graph of y=2x+6. A coordinate system has a horizontal x-axis labeled from negative 5 to 5 in increments of 1 and a vertical y-axis labeled from negative 20 to 20 in increments of 1. A line rises from left to right and passes through the plotted points (negative 3, 0), (negative 2, 2), (negative 1, 4), (0, 6), (1, 8), (2, 10), and (3, 12).

1.2 Determine whether the following equation defines y as a function of x. lxl−y=3 Does the equation lxl−y=3 define y as a function of​ x? Yes No

Yes

1.2 Determine whether the following equation defines y as a function of x. xy+3y=3 Does the equation xy+3y=3 define y as a function of​ x? Yes No

Yes

1.2 Determine whether the following equation defines y as a function of x. x^ 2+y=1 Does the equation x^2+y=1 define y as a function of​ x? Yes No

Yes

1.4 Use the given conditions to write an equation for the line in​ point-slope form and​ slope-intercept form. Slope=−8​, passing through (−5,0) a) Type the​ point-slope form of the line. ______ ​(Simplify your answer. Use integers or fractions for any numbers in the​ equation.) b) Type the​ slope-intercept form of the line. _______ ​(Simplify your answer. Use integers or fractions for any numbers in the​ equation.)

a) ​ y−0=−8(x+5) b)y=−8x−40

1.1 Use the graph and a. determine the​ x-intercepts, if​ any; b. determine the​ y-intercepts, if any.

a. An​ x-intercept of a graph is the​ x-coordinate of a point where the graph intersects the​ x-axis. The point where the given graph intersects the ​x-axis,(−2, 0)​, has an​ x-coordinate of −2. Since the graph intersects the​ x-axis at the point (−2, 0), the​ x-intercept is −2. b. A​ y-intercept of a graph is the​ y-coordinate of a point where the graph intersects the​ y-axis. Since the graph does not intersect the​ y-axis at any​ point, there is no​ y-intercept.

1.1 Use the graph to the right to complete the following. For the​ graph, tick marks along the axes represent one unit each. a. Determine the​ x-intercepts, if any. b. Determine the​ y-intercepts, if any.

a. The​ x-intercept is the​ x-coordinate of a point where the graph intersects the​ x-axis. ​First, determine the number of times the graph intersects the​ x-axis, if any. Notice that the graph intersects the​ x-axis 2 times. Now determine the points at which the graph intersects the​ x-axis. Notice that dots have been drawn showing the points where the graph intersects the​ x-axis. The graph intersects the​ x-axis at ​(−3​,0) and ​(1​,0). Using the results from the previous​ step, the​ x-intercepts are −3 and 1. b. The​ y-intercept is the​ y-coordinate of a point where the graph intersects the​ y-axis. ​First, determine the number of times the graph intersects the​ y-axis, if any. Notice that the graph intersects the​ y-axis 1 time. Now determine the point at which the graph intersects the​ y-axis. A dot has been drawn showing the point where the graph intersects the​ y-axis. The graph intersects the​ y-axis at ​(0,−9​). Using the results from the previous​ step, the​ y-intercept is −9.

1.2 Use the graph to determine the following. a. the​ function's domain b. the​ function's range c. the​ x-intercepts, if any d. the​ y-intercept , if any e. the function value indicated below. f(2) A coordinate system has a horizontal x-axis labeled from negative 5 to 5 in increments of 1 and vertical y-axis labeled from negative 5 to 5 in increments of 1. A curve falls from left to right from a closed circle at (0, 0) to an open circle at (4, negative 4) passing through the point (2, negative 3).

a. To determine the domain of a function given its​ graph, look for all the inputs on the ​ x-axis that correspond to points on the graph. The domain will be the largest set of real numbers represented by the​ x-coordinates. Notice that the points on the graph have​ x-coordinates that extend from 0 to 4. The closed dot at the point ​(0​,0​) indicates that 0 is included in the domain. The open dot at the point ​(4​,−4​) indicates that 4 is not included in the domain.​ Therefore, the domain of f in interval notation is [0,4). b. To determine the range of a function given its​ graph, examine the​ y-coordinates on the graph of the function. The range will be the set of all real numbers represented by the​ y-coordinates. Notice that the points on the given graph of f have​ y-coordinates that extend from 0 to −4. The closed dot at the point ​(0​,0​) indicates that 0 is included in the range. The open dot at the point ​(4​,−4​) indicates that −4 is not included in the range.​ Therefore, the range of f in interval notation is (−4,0]. c. On the graph of a​ function, x-intercepts occur where the graph touches or crosses the​ x-axis. At those​ points, the​ y-coordinates are equal to 0. The graph of f shown touches the​ x-axis at the point ​(0​,0​). ​ Therefore, the​ x-intercept is 0. d. On the graph of a​ function, the​ y-intercept occurs where the graph touches or crosses the​ y-axis. At that​ point, the​ x-coordinate is equal to 0. The graph of f shown touches the​ y-axis at the point ​(0,0​). ​Therefore, the​ y-intercept is 0. e. In order to find the specific value of a function from a given​ graph, we need to locate the point on the graph having the specific​ x-coordinate and determine the​ y-coordinate of that ordered pair. To find ​f(2​), which is the value of the given function when x​ = 2​, locate the point on the graph of the function that has the​ x-coordinate 2. Notice that the​ y-coordinate of this point is −3. ​ Therefore, f(2)=−3.

1.2 Use the graph to determine a. the​ function's domain; b. the​ function's range; c. the​ x-intercepts, if​ any; d. the​ y-intercept, if​ any; and e. the missing function​ value, indicated by the question​ mark, below. f(2)=​?

a. To find the​ domain, look for all the inputs on the​ x-axis that correspond to points on the graph. Notice that the points on the graph have​ x-coordinates that extend from −∞ to ∞. The domain is (−∞,∞). b. To find the​ range, look for all the outputs on the​ y-axis that correspond to points on the graph. Notice that the points on the given graph of f have​ y-coordinates that extend from 0 to ∞, but not including 0. There are no points on the graph of f below the line y=0. The range is (0,∞). c. On the graph of a​ function, x-intercepts occur where the graph touches or crosses the​ x-axis. At those​ points, the​ y-coordinates are equal to 0. The graph shown above approaches but never touches the​ x-axis. Therefore, there is no​ x-intercept. d. On the graph of a​ function, the​ y-intercept occurs where the graph touches or crosses the​ y-axis. At that​ point, the​ x-coordinate is equal to 0. The​ y-intercept is 4. e. In order to find a specific value of a function from a given​ graph, locate the point on the graph having the specified​ x-coordinate and determine the​ y-coordinate of that ordered pair. Use this procedure to find the indicated function value. f(2)=2

1.2 Use the graph to determine a. the​ function's domain; b. the​ function's range; c. the​ x-intercepts, if​ any; d. the​ y-intercept, if​ any; and e. the missing function​ value, indicated by the question mark below. f(4)=​?

a. To find the​ domain, look for all the inputs on the​ x-axis that correspond to points on the graph. The left boundary for the​ x-coordinates of the points of the graph is 0. There is no right boundary for the​ x-coordinates of the points of the graph. There are no gaps or values of x that do not occur to the right of x=0. Since the lowest value of x for points on the graph is​ 0, and there is no highest value of x and no​ gaps, the values extend from 0 to ∞ and include 0.​ Therefore, the domain is [0, ∞). b. To find the​ range, look for all the outputs on the​ y-axis that correspond to points on the graph. Notice that the points on the given graph of f have​ y-coordinates that extend from 10 to ∞ and include 10. There are no points on the graph of f below the​ y-coordinate 10. ​Therefore, the range is [10, ∞). c. On the graph of a​ function, x-intercepts occur where the​ y-coordinates are equal to 0. At such​ points, the graph touches or crosses the​ x-axis. Therefore, there are no​ x-intercepts. d. On the graph of a​ function, the​ y-intercept occurs where the​ x-coordinate is equal to 0. At this​ point, the graph touches or crosses the​ y-axis. Therefore, the​ y-intercept is 10. e. To find a specific value of a function from a given​ graph, locate the point on the graph having the specified​ x-coordinate and determine the​ y-coordinate of that ordered pair. Use this procedure to find the indicated function value. f(4)=12

1.2 Use the graph to determine a. the​ function's domain; b. the​ function's range; c. the​ x-intercepts, if​ any; d. the​ y-intercept, if​ any; and e. the missing function​ values, indicated by question​ marks, below. f(−3)=​? f(−1)=​?

a. To find the​ domain, look for all the inputs on the​ x-axis that correspond to points on the graph. There is no left boundary for the​ x-coordinates of the points of the graph. There is no right boundary for the​ x-coordinates of the points of the graph. There are no​ gaps, or values of x that do not occur. Notice that the points on the graph have no lowest value for their​ x-coordinates, and no highest​ value, and no gaps. The values extend from −∞ to ∞. ​Therefore, the domain of f is (−∞,∞). b. To find the​ range, look for all the outputs on the​ y-axis that correspond to points on the graph. Notice that the points on the given graph of f have​ y-coordinates that extend from −∞ up to​ 4, and includes 4. There are no points on the graph of f above the​ y-coordinate 4.​ Therefore, the range is (−∞,4]. c. On the graph of a​ function, x-intercepts occur where the​ y-coordinates are equal to 0. At such​ points, the graph touches or crosses the​ x-axis. The​ x-intercepts are −2 and 2. d. On the graph of a​ function, the​ y-intercept occurs where the​ x-coordinate is equal to 0. At this​ point, the graph touches or crosses the​ y-axis. The​ y-intercept is 4. e. In order to find the specific value of a function from a given​ graph, locate the point on the graph having the specific​ x-coordinate and determine the​ y-coordinate of that ordered pair. Use this procedure to find the indicated function values. f(−3) = −5 f(−1) = 3