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

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!
Screenshot_5.png30.8KB
Reply 1
Avatar for gomc
gomc
11
it means that when x =0, y=4.
Well sub x=0 into y=x^2+5x+4.
Reply 3
Avatar for Tau
Tau
10
The point at which the graph crosses the y-axis is the point at which x=0 x = 0 . If you substitute x=0 x = 0 into x2+5x+4 x^2 + 5x + 4 , the answer is 4.
Reply 4
Avatar for blobbybill
blobbybill
OP
11
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?
Reply 5
Avatar for Tau
Tau
10
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 x = 0 into a quadratic y=ax2+bx+c y = ax^2 + bx + c the y coordinate will always be c c
Reply 6
Avatar for blobbybill
blobbybill
OP
11
Original post by Tau
Essentially, yes. Note though that if you substitute x=0 x = 0 into a quadratic y=ax2+bx+c y = ax^2 + bx + c the y coordinate will always be c 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.


Screenshot_7.png
Reply 7
Avatar for Tau
Tau
10
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=ax2+bx+c y = ax^2 + bx +c or in your case, y=x2+5x+4 y = x^2 + 5x + 4 . If y=0 y = 0 then x2+5x+4=0 x^2 + 5x +4 = 0 , so (x+1)(x+4)=0 (x+1)(x+4) = 0 .
Reply 8
Avatar for blobbybill
blobbybill
OP
11
Original post by Tau
You have an equation of the form y=ax2+bx+c y = ax^2 + bx +c or in your case, y=x2+5x+4 y = x^2 + 5x + 4 . If y=0 y = 0 then x2+5x+4=0 x^2 + 5x +4 = 0 , so (x+1)(x+4)=0 (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?
Reply 9
Avatar for Tau
Tau
10
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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