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Can someone help with Quadratic Graphs please?

Maths homework please help quickly!, So confused?

  • Hi so on my maths homework it says 'draw the graphs of these straight lines for the given range of values of x' a. y + 4x + 8 = 0 for x = -2 to x = 5 b. y - 2x - 5 = 0 for x = -4 to 2 Can someone explain to me what to do please? I asked my dad as well and he doesn't seem to understand it either, so I really really REALLY need help. Thank you so much in advance :)

  • Answer:

    let x = -2, y+(4*-2)+8=0 y-8+8=0 y=0.....................so (-2,0) plot this point letx=5 y+(4*5)+8=0 y+20+8=0 y+28=0 y=-28................so(5,-28) plot this point...and join the two points together b) let x=-4 y-(2*-4)-5=0 y-(-8)-5=0 y+8-5=0 y+3=0 y=-3...............(-4,-3) plot this point let x=2 y-(2*2)-5=0 y-(4)-5=0 y-4-5=0 y-9=0 y=9...............(2,9) plot this point and join the two point together

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this really helps to plot graphs and see where you're going wrong http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html

let x = -2, y+(4*-2)+8=0 y-8+8=0 y=0.....................so (-2,0) plot this point letx=5 y+(4*5)+8=0 y+20+8=0 y+28=0 y=-28................so(5,-28) plot this point...and join the two points together b) let x=-4 y-(2*-4)-5=0 y-(-8)-5=0 y+8-5=0 y+3=0 y=-3...............(-4,-3) plot this point let x=2 y-(2*2)-5=0 y-(4)-5=0 y-4-5=0 y-9=0 y=9...............(2,9) plot this point and join the two point together

Mark

Let's start with question A... Draw the graphs of these straight lines for the given range of values of x: a. y + 4x + 8 = 0 for x = -2 to x = 5 First, calculate where Y would be if x = -2 y + 4x + 8 = 0 y + 4(-2) + 8 = 0 y - 8 + 8 = 0 y = 0 So now you know that ONE of the plot points on your line is ( - 2, 0 ) Since the line is straight, you do not need to do a series of plotted points, just one more point to define the line. So, calculate where Y would be if x = 5 y + 4x + 8 = 0 y + 4(5) + 8 = 0 y + 20 + 8 = 0 y + 28 = 0 y = -28 So the another plot point is ( 5 , -28 ). Now plot those points on a graph, and draw a straight line (infinitely long), using it. Do (b) second one in a similar fashion.

Nathan Birnbaum

the equations are for straight lines you just need to find the points to connect a. y + 4x + 8 = 0 for x = -2 to x = 5 when x= -2, y=0 when x=5, y= -28 plot (-2,0) and (5, -28) connect the points b. y - 2x - 5 = 0 for x = -4 to 2 when x= -4, y=13 when x=2, y=9 plot (-4,13) and (2,9) connect the points

Steven

a) substitute x = -2 and x = 5 into the equation and calculate the corresponding values of y. You'll have two points (x, y). Plot those points and draw a straight line between them. b) do the same thing using x = -4 and x = 2 in the second equation. ...

TomV

Let's start with question A... Draw the graphs of these straight lines for the given range of values of x: a. y + 4x + 8 = 0 for x = -2 to x = 5 First, calculate where Y would be if x = -2 y + 4x + 8 = 0 y + 4(-2) + 8 = 0 y - 8 + 8 = 0 y = 0 So now you know that ONE of the plot points on your line is ( - 2, 0 ) Since the line is straight, you do not need to do a series of plotted points, just one more point to define the line. So, calculate where Y would be if x = 5 y + 4x + 8 = 0 y + 4(5) + 8 = 0 y + 20 + 8 = 0 y + 28 = 0 y = -28 So the another plot point is ( 5 , -28 ). Now plot those points on a graph, and draw a straight line (infinitely long), using it. Do (b) second one in a similar fashion.

Nathan Birnbaum

a) substitute x = -2 and x = 5 into the equation and calculate the corresponding values of y. You'll have two points (x, y). Plot those points and draw a straight line between them. b) do the same thing using x = -4 and x = 2 in the second equation. ...

TomV

the equations are for straight lines you just need to find the points to connect a. y + 4x + 8 = 0 for x = -2 to x = 5 when x= -2, y=0 when x=5, y= -28 plot (-2,0) and (5, -28) connect the points b. y - 2x - 5 = 0 for x = -4 to 2 when x= -4, y=13 when x=2, y=9 plot (-4,13) and (2,9) connect the points

Steven

this really helps to plot graphs and see where you're going wrong http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html

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