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Math Help - I have Slopes And Points , How do i graph it?

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    I have Slopes And Points , How do i graph it?

    hi, I have the m which is the slope and 2 random x and y points.
    For example now i have

    m= -3/2 and two points i got are (4,8)
    And i have been asked to make a graph out of this. How would i do this?
    Thanks For all the help.
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    Quote Originally Posted by satishgaire View Post
    hi, I have the m which is the slope and 2 random x and y points.
    For example now i have

    m= -3/2 and two points i got are (4,8)
    And i have been asked to make a graph out of this. How would i do this?
    Thanks For all the help.
    i suppose you want to graph a straight line? if you get the equation of the line, could you take it from there?

    here's how we get the equation:


    method 1: use the point-slope form and solve for y.

    the equation of a line with slope m passing through a point (x_1,y_1) is given by

    y - y_1 = m(x - x_1)

    plug in your values and solve for y


    method 2: use the slope-intercept form, plug in your knowns and solve for the unknown.

    recall that the equation of a line is y = mx + b, where m is the slope and b is the y-intercept

    you know m and you know and x value and its corresponding y-value. so plug those in and solve for b. then you can write the equation of the line


    can you continue?
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    so, it would be y=-3/2x+b
    but i have 2 points 4,8 and in which 4 is x and 8 is y
    and as b= Y intercept
    So, my equation would be y=-3/2x+8
    Now i can simple put this in my TI Calculator and then graph it.

    Am i correct?

    Also. I see this topic got moved to Pre-Calculus. and i am in algebra class
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    Quote Originally Posted by satishgaire View Post
    so, it would be y=-3/2x+b
    but i have 2 points 4,8 and in which 4 is x and 8 is y
    and as b= Y intercept
    So, my equation would be y=-3/2x+8
    Now i can simple put this in my TI Calculator and then graph it.

    Am i correct?

    Also. I see this topic got moved to Pre-Calculus. and i am in algebra class
    no. please re-read my first post. it seems you missed something

    your equation is 8 = (-3/2)*4 + b

    there is no way b is 8. (there is a difference between y-value and y-intercept! y-intercept refers to a specific y-value, namely the y-value when x = 0. 8 is the y-value when x = 4, and hence, it is not the y-intercept)

    you're supposed to graph with your TI calculator?
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    They never said don't do it with my TI Calculator. We are allowed to use this calculator on Texas's State based Tests and other regular tests and quizes .

    So.

    y=mx+b
    8=-3/2(4)+b
    8=-6+b
    +6 to the negetive 6 so that cancels out.
    8= b
    +6 add the 6 to 8 and that becomes 14

    14= B

    So my equation is

    y=-3/2x+14

    is that the correct equation?
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    Quote Originally Posted by satishgaire View Post
    They never said don't do it with my TI Calculator. We are allowed to use this calculator on Texas's State based Tests and other regular tests and quizes .

    So.

    y=mx+b
    8=-3/2(4)+b
    8=-6+b
    +6 to the negetive 6 so that cancels out.
    8= b
    +6 add the 6 to 8 and that becomes 14

    14= B

    So my equation is

    y=-3/2x+14

    is that the correct equation?
    yes, that is correct.

    try graphing it without your calculator first. you can use the calculator to check it.

    to graph a straight line, all you need are two points. then you draw a straight line through them.

    we already have the point (4,8), so we can draw our axis and plot that point. now we need one other point. the easiest points to find are the x- and y-intercepts. the easier one is the y-intercept, since we know it already. it is (0,14) (remember, the x-value is always 0 for the y-intercept, so we don't have to calculate anything). so that is the second point you should plot. then draw a straight line through both points
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    Thanks for that but what if the m which is slope is undefined. and i just have the 2 points. how would i be getting that equation to graph it?

    thans
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    Quote Originally Posted by satishgaire View Post
    Thanks for that but what if the m which is slope is undefined. and i just have the 2 points. how would i be getting that equation to graph it?

    thans
    if m is undefined, you have a vertical line. these are of the form x = (some constant). so they have their own rules. as long as you know one point on the line, you will know the line, because it is just x = (whatever x-value you were given)
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    ok line is vertical but where will that vertical line be on the graph.
    for example the points i got are -9,6 so i will go to x intercept and where the 6 x is i would make a vertical line on there? i doubt that is correct.
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    Quote Originally Posted by satishgaire View Post
    ok line is vertical but where will that vertical line be on the graph.
    for example the points i got are -9,6 so i will go to x intercept and where the 6 x is i would make a vertical line on there? i doubt that is correct.
    yes, just draw a vertical line passing through -9 on the x-axis
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