Value of b, y-intercept of Quadratic graph

[gazparkin]

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!

 

[Greg]

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

 

[HallsofIvy]

The graph is of $$y= x^2+ ax+ b$$ and we are told that the graph goes through (2, 0). That means that when x= 2, y= 0. So we must have $$0= 2^2+ a(2)+ b= 4+ 2a+ b$$ 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 $$0= 4^2+ a(4)+ b= 16+ 4a+ b$$ or 4a+ b= -16.

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

 

[gazparkin]

The graph is of $$y= x^2+ ax+ b$$ and we are told that the graph goes through (2, 0). That means that when x= 2, y= 0. So we must have $$0= 2^2+ a(2)+ b= 4+ 2a+ b$$ 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 $$0= 4^2+ a(4)+ b= 16+ 4a+ b$$ 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.