Question 478378: If possible there are several questions that are puzzling me and if possible I would like some help. Graphing is very confusing and I really need all the help I can get...thank you in advance
1.Graphing f(x) = 4x-1
2.Graphing g(x)=|x+2|
3. Use the graph of the function f to find f(-4),f(-3),f(-2)
4. Graph f(x)= -3x
5. graph the function f(x)=x²-x-5
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 1.Graphing f(x) = 4x-1
2.Graphing g(x)=|x+2|
3. Use the graph of the function f to find f(-4),f(-3),f(-2)
4. Graph f(x)= -3x
5. graph the function f(x)=x²-x-5
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1.Graphing f(x) = 4x-1
You might want to note that if x is of the first degree, the graph will be a straight line, so you need only two points to graph it. The standard form of an equation for a straight line is y=mx+b, with m=slope, and b=y-intercept (where line crosses the y-axis).
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For given equation, y=4x-1, the slope m=4 and the y-intercept=-1, which gives you one point.
Use the slope =4, to find the second point. From the y-intercept of -1, go 4 units up and 1 unit to the right. (rise/run). This is your second point which can connect to the first point to draw a line.
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2.Graphing g(x)=|x+2|
The absolute value of |x+2| means that y can never be negative. You should familiarise yourself with the function y=|x| which is a v-shaped curve with the bottom apex at (0,0).
y=|x+2| just means shifting this v-shaped curve 2 units left.
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3. Use the graph of the function f to find f(-4),f(-3),f(-2)
f(x)=4x-1
f(-4)=-16-1=-17
f(-3)=-12-1=-13
f(-2)=-8-1=-9
This means points (-4,-17), (-3,-13) and (-2,-9) are on the straight line of f
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4. Graph f(x)= -3x
This does not look like an equation for the straight line, y=mx+b, but it is. The y-intercept, b does not appear because it is zero, that is, it goes thru the origin (0,0). Again you can find a second point starting from the origin and going 3 units down and 1 unit right to give you a line with a negative slope of 3.
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5. graph the function f(x)=x²-x-5
This function has x of a second degree, so it is not a straight line, but a parabola. To graph a parabola we need to see the equation in standard form: y=A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex.
y=x^2-x-5
completing the square
y=(x^2-x+1/4)-5-1/4
y=(x-1/2)^2-20/4-1/4
y=(x-1/2)^2-21/4
vertex: (1/2,-21/4) (This gives you one point on the curve.)
You can get a second point by finding the y-intercept. Setting x=0, y=-5
So a second point is (0,-5). Since the curve is symmetrical around its axis of symmetry (x=1/2), we can easily find the third point to be (1,-5)
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Note: You can always plot points to graph a function, but I hope I gave you some ideas which would be faster and more effective. If you are really interested in learning how to graph, I suggest you learn how to use a graphing calculator or a graphing computer program. Anyway, I hope I helped.
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