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Value of b, y-intercept of Quadratic graph

gazparkin

New member
Jul 1, 2019
17
Hi,

Can anyone help me understand how I get to the answer on this one?

The diagram shows a sketch of the graph of y = x2 + ax + b

The graph crosses the x-axis at (2, 0) and (4, 0).

Work out the value of b.


Thank you in advance!
 

Attachments

Greg

Perseverance
Staff member
\(\displaystyle y=(x - 2)(x - 4)=x^2-6x+8\)
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
The graph is of [tex]y= x^2+ ax+ b[/tex] and we are told that the graph goes through (2, 0). That means that when x= 2, y= 0. So we must have [tex]0= 2^2+ a(2)+ b= 4+ 2a+ b[/tex] or 2a+ b= -4. We are also told that the graph goes through (4, 0). That means that when x= 4, y= 0. So we must have [tex]0= 4^2+ a(4)+ b= 16+ 4a+ b[/tex] or 4a+ b= -16.

Solve the two equations, 2a+ b= -4 and 4a+ b= -16, for a and b.
 

gazparkin

New member
Jul 1, 2019
17
The graph is of [tex]y= x^2+ ax+ b[/tex] and we are told that the graph goes through (2, 0). That means that when x= 2, y= 0. So we must have [tex]0= 2^2+ a(2)+ b= 4+ 2a+ b[/tex] or 2a+ b= -4. We are also told that the graph goes through (4, 0). That means that when x= 4, y= 0. So we must have [tex]0= 4^2+ a(4)+ b= 16+ 4a+ b[/tex] or 4a+ b= -16.

Solve the two equations, 2a+ b= -4 and 4a+ b= -16, for a and b.
Thank you for this - really helped me understand.