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# how are functions y = x and y-3 related and how are their graphs related

need to know how functions are related.

## Research, Knowledge and Information :

### How are the functions y = x and y = x – 3 related? How are ...

How are the functions y = x and y = x – 3 related? ... How are the functions y = x and y = x – 3 related? How are their graphs related? . Can someone explain this ...

### How are the functions y = x and y = x – 3 related... - OpenStudy

How are the functions y = x and y = x – 3 related? How are their graphs related? ... How are the functions y = x and y = x – 3 related? How are their graphs related?

### describe how the graphes of y=/x/ and y=/x/-15 are related A ...

... x-2) 2.Describe how the graph of y=/x/ and y=/x+3/ are related. A.The two graphs ... x| and y=|x|-5. how are the functions related? ... their graphs in one ...

### Functions and Their Graphs - Math24

Functions and Their Graphs. ... (where $$\mathbb{R}$$ is the set of real numbers), then the function $$y = f\left( x \right)$$ ... Related Pages.

### Functions and Their Graphs - NIU - Northern Illinois University

Chapter 3 Functions and Their Graphs ... There are also some basic models related to ... FUNCTIONS AND THEIR GRAPHS 3.2 #25c. For the function f(x) = x+2 x 6 ...

### How are the functions y = x and y = x – 3 related? How are ...

How are the functions y = x and y = x – 3 related? How are their graphs ... x – 3 related? How are their graphs related? ... y=3 and x=3. Are those relations ...

### Graph of a function - Wikipedia

The graph of a function on real numbers may be mapped directly to ... and other areas, graphs are tools used ... Wikimedia Commons has media related to Function ...

## Suggested Questions And Answer :

### he graph of the function has one relative maximum and relative minimum point.

f(x)=x^3-12x=x(x^2-12), and f'(x)=3x^2-12=0 at extrema: x^2=4 so x=+2. f(2)=-16, f(-2)=16, so the points are (2,-16) and (-2,16). The graph passes through zero, implying that one turning point is maximum and the other minimum. f(1)=-11 and f(-1)=11, so we can deduce that (2,-16) is a minimum and (-2,16) is a maximum because f(x) in the vicinity has a value nearer to zero then the turning point. f"(x)=6x, f"(2)>0 (minimum), f"(-2)<0 (maximum).

### can you help me before saturday ?

a). First put the equations in terms you are used too. f(x) = 1.25x + 20 g(x) = 1.25x + 10 they are linear (straight lines) and they are 10 units apart. f(x) goes through (0, 20) and g(x) goes through (0, 10) f(1) = 1.25 + 20 = 21.25 (1,21.25) g(1) = 1.25 + 10 = 11.25 (1, 11.25) There are the 2 point for each function to graph them. b). The lines are parellel. c). make the constant equal.  (20 and 10) d). the function would now be F(x) >= 1.25(x + 150) F(x) >= 1.25x + 187.50 It would be the line and all the values above it It would start at (0, 187.50)

### Using table or graph write a linear function that relates y to x.

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### How do you set up this word problem?

After visiting relatives who live 200 miles away, your family drives home at an average speed of 50 miles per hour. Your distance d( in miles) from home is given by d=200-50t where t is the time (in hours)spent driving. Graph the function and identify its domain and range. What is your distance from home after driving for 1.5 hours? This is a linear equation and the values are decreasing. Start at the point (0, 200) on the y-axis go down 50 and over 1 this gives you the line.  domain is {0, 2} hours and range is {0 to 200} miles. d = 200 - 50t at t = 1.5 plug in d = 200 - 50(1.5) d = 200 - 75 d = 125 mi

### how do you graph relations and functions?

the values of x for which 4x2 - 12x - (-3)=0 expressing your answers to three significant figures

### What is relation and function

funkshun...y=f(x) . . . graf is seeabel wae tu sho funkshun . . . relashun...kin folk

### find the exact location OF ALL THE absolute and relative extrema of the function f(x)=x^3-6x^2 +9x+1

The extreme are found by differentiating f(x): 3x^2-12x+9 and solving for f'(x)=0. Divide f'(x)=0 through by 3: (x-1)(x-3)=0. f(1)=5; f(3)=1. Differentiating again: f''(x)=6(x-2). f''(1)=-6 and f''(3)=6, therefore (1,5) is a maximum, and (3,1) is a minimum. The graph shows f(x)<0 when x<-0.104 (approx), crosses the vertical axis at (0,1), reaches the maximum at (1,5) dipping down to the minimum at (3,1), and rising thereafter (f(x)>0). For x>-0.104 f(x) remains positive.

### integral from 0 to infinity of (cos x * cos(x^2)) dx

The behaviour of this function f(x)=cos(x)cos(x^2) is interesting. The integral is the area between the curve and the x axis. If the functions cos(x) and -cos(x) are plotted on the same graph, the latter form an envelope for f(x). Between x=(pi)/2 and 3(pi)/2, the curve has 4 maxima and 3 minima; between x=3(pi)/2 and 5(pi)/2 there are 7 maxima and 6 minima; between x=(2n-1)(pi)/2 and (2n+1)(pi)/2 there are 3n+1 maxima and 3n minima (integer n>0), a total of 6n+1. (These are based purely on observation, and need to be supported by sound mathematical deduction.) As n becomes large the envelope appears to fill as the extrema become closer together. As x tends to infinity n also tends to infinity. The envelope apparently has as much area above (positive) as below (negative) the x axis so the total area will be zero as the positive and negative areas cancel out. The question is: do the areas cancel out exactly? As x gets larger, the curve starts to develop irregularities and patterns, but it stays within the envelope, and positive irregularities appear to be balanced by negative irregularities, so the overall symmetry appears to be preserved. f(x)=0 when cos(x)=0 or cos(x^2)=0, which means that x=(2n-1)(pi)/2 or sqrt((2n-1)(pi)/2), where n>0. Between 3(pi)/2 and 5(pi)/2, for example, we have sqrt(3(pi)/2), sqrt(5(pi)/2), ..., sqrt(13(pi)/2), because sqrt(13(pi)/2)<3(pi)/20, and this lies between (2n-1) and (2n+1); so n is defined by 2n-1<(2m+1)^2(pi)/2<2n+1.  For m-1 we have 2(n-z)-1<(2m-1)^2(pi)/2<2(n-z)+1, where z is related to the number of zeroes in the current "batch". For example, take m=3: 2n-1<49(pi)/2<2n+1; 49(pi)/2=76.97 approx., so 2n-1=75, and n=38. Also 2(n-z)-1<25(pi)/2<2(n-z)+1 so, because 25(pi)/2=39.27 approx., 2(n-z)+1=41, n-z=20, and z=18. When m=2, 2n-1=39, n=20; 2(20-z)-1<9(pi)/2<2(20-z)+1; 2(20-z)-1=13, 20-z=7, z=13. The actual number of zeroes, Z, including the end points is 2 more than this: Z=z+2. Now we have an exact way to calculate the number of zeroes in each batch. So Z and n are both related to m. The number of extrema, E=Z-1=z+1. In fact, E=int(2(pi)(m-1)+1), where int(a) means the integer part of a, so as m increases, there are proportionately more extrema over the range (2m-1)(pi)/2 to (2m+1)(pi)/2. The figure of 6n+1 deduced earlier by observation approximates to the mathematical findings, because 2(pi) is approximately equal to 6. But we still need to show, or disprove, that the areas above and below the x axis are equal and therefore cancel out. Unfortunately, if we consider the area under the first maximum (between x=(pi)/2 and sqrt(3(pi)/2)), and the area above the first minimum (between x=sqrt(3(pi)/2) and sqrt(5(pi)/2)), they are not the same, so do not cancel out. More...

### what is a function

a function is a relation between x and y, where an x can't go to two different y's on a graph of f(x), no point can be directly above or below each other this is known as the vertical line test: a vertical line cannot pass through two different points of the function a circle is not a function y=ax^2 +bx +c, a parabola the has a U shape, or upside down U shape is a function

### cast rule behavior

360 degrees is divided into 4 quadrants: Q1: 0-90; Q2=90-180: Q3=180-270; Q4=270-360. In Q1, all trig functions are positive; in Q2, only sine is positive (sin(X)=sin(180-X)); in Q3, only tangent is positive and in Q4 only cosine is positive. CAST=cosine,all,sine,tangent=Q4,Q1,Q2,Q3. Graphically cosine and sine look similar but they are displaced by a phase difference of 90 degrees. Tangent resembles neither cosine nor sine, but is nevertheless periodic in that the pattern repeats. The red curve is y=sin(x), blue is y=cos)x), green is y=tan(x) (asymptotes are shown as green vertical lines). Between the y-axis and the first green line to the right all functions are positive (Q1), where the red and green curves intersect further to the right, we have Q2, Q3 is where the green curve is positive, up to the next vertical green line, and Q4 goes off the picture, but Q4 is also up to the first green vertical line to the left of the y-axis, followed further left by Q3, etc. The regular pattern continues indefinitely repeating every 360 degrees. The gap between the green verticals is 180 degrees.