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C1 Sketching Quadratic solution, where does the bottom line come from? Watch

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    We haven't been taught how to do it, but it is for our homework. Ihave gone on the solution bank to try and work out how to do it, and this is the answer and solution (image attached).

    I can follow and understand it up to where it says the x axis crossing points are (-1,0) and (-4,0) because they are the values of x when you solve it via factorising.
    However, I cannot work out for the life of me where the whole bottom line comes from, and what it means when it says: x=0, y=4 so y axis crossing point is (0,4). Can you please explain where the x=0 and y=4 comes from please?

    As I said, I haven't been taught how to do it (we have to teach ourselves this bit for homework) and I can't understand it.
    Thanks!
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    it means that when x =0, y=4.
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    Well sub x=0 into y=x^2+5x+4.
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    The point at which the graph crosses the y-axis is the point at which  x = 0 . If you substitute  x = 0 into  x^2 + 5x + 4 , the answer is 4.
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    (Original post by an_atheist)
    Well sub x=0 into y=x^2+5x+4.
    When solving a question like this though, would you always do it exactly the same way as this solution? By that, I mean would you work out both x values by factorising (like they did in this solution), and then always sub x=0 into work out the y-axis point?Like do you always sub y=0 when you work out the x axis value, and then you also sub x=0 in to work out the y coordinate?
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    (Original post by blobbybill)
    When solving a question like this though, would you always do it exactly the same way as this solution? By that, I mean would you work out both x values by factorising (like they did in this solution), and then always sub x=0 into work out the y-axis point?Like do you always sub y=0 when you work out the x axis value, and then you also sub x=0 in to work out the y coordinate?
    Essentially, yes. Note though that if you substitute  x = 0 into a quadratic  y = ax^2 + bx + c the y coordinate will always be  c
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    (Original post by Tau)
    Essentially, yes. Note though that if you substitute  x = 0 into a quadratic  y = ax^2 + bx + c the y coordinate will always be  c
    Thanks. I have forgotten what this highlighted bit comes from, what it means and why you do it. I get that you sub y=0 to work out the x coordinates, but how do you know what you factorise into in order to get (x+1)(x+4)? What equation does that derive from?

    ie, how do you know what to factorise, in order to get (x+1)(x+4)??. Sorry, I have just gotten really confused.


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    (Original post by blobbybill)
    Thanks. I have forgotten what this highlighted bit comes from, what it means and why you do it. I get that you sub y=0 to work out the x coordinates, but how do you know what you factorise into in order to get (x+1)(x+4)? What equation does that derive from?

    ie, how do you know what to factorise, in order to get (x+1)(x+4)??. Sorry, I have just gotten really confused.
    You have an equation of the form  y = ax^2 + bx +c or in your case,  y = x^2 + 5x + 4 . If  y = 0 then  x^2 + 5x +4 = 0 , so  (x+1)(x+4) = 0 .
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    (Original post by Tau)
    You have an equation of the form  y = ax^2 + bx +c or in your case,  y = x^2 + 5x + 4 . If  y = 0 then  x^2 + 5x +4 = 0 , so  (x+1)(x+4) = 0 .
    So to find the x coordinates you sub y=0 into it and factorise it to get the values of x, and to find the y coordinate you sub x=0 into the equation?
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    (Original post by blobbybill)
    So to find the x coordinates you sub y=0 into it and factorise it to get the values of x, and to find the y coordinate you sub x=0 into the equation?
    Yes, although when you substitute y=0 you won't always be able to factorise it, but you just solve the quadratic.
 
 
 
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