Value of b, y-intercept of Quadratic graph

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SUMMARY

The discussion focuses on determining the value of b in the quadratic equation y = x² + ax + b, given that the graph intersects the x-axis at points (2, 0) and (4, 0). The factored form of the quadratic is y = (x - 2)(x - 4), which expands to y = x² - 6x + 8. By substituting the x-values into the equation, two linear equations are formed: 2a + b = -4 and 4a + b = -16. Solving these equations yields the values of a and b, confirming that b = 8.

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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.
 

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