MHB Value of b, y-intercept of Quadratic graph

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To find the value of b in the quadratic equation y = x^2 + ax + b, given that the graph crosses the x-axis at (2, 0) and (4, 0), we can derive two equations from these points. Substituting x = 2 into the equation gives 2a + b = -4, and substituting x = 4 results in 4a + b = -16. By solving these simultaneous equations, we can isolate a and b. The solution reveals the value of b as 8. Understanding this process clarifies the relationship between the coefficients and the graph's intercepts.
gazparkin
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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!
 

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$$y=(x - 2)(x - 4)=x^2-6x+8$$
 
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.
 
HallsofIvy said:
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.
 

Thread 'Area of Overlapping Squares'

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