MHB Find equation of a quadratic with line of symmetry

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The axis of symmetry for a quadratic equation is given by x = -b/(2a). If an x-intercept is at (-2, 0), it leads to the equation 0 = a(-2)^2 + b(-2) + c, establishing relationships between b and c as functions of a. When the axis of symmetry is x = 1.75, the quadratic can be expressed as y = a(x - 1.75)^2 + c, resulting in b = 3.5a. Additionally, with the condition y = 0 when x = -2, it simplifies to 0 = 4a - 2b + 22, yielding b = 11 + 2a. This discussion highlights the interdependence of coefficients in forming a quadratic equation based on its symmetry and intercepts.
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Here's something to start you off. The axis of symmetry for the parabola [math]y = ax^2 + bx + c[/math] is [math]x = -\dfrac{b}{2a}[/math] and if an x-intercept is at (-2, 0) then [math]0 = a(-2)^2 + b(-2) + c[/math]. That will give you both b and c as functions of a.

Can you finish?

-Dan
 
Equivalently, if the axis of symmetry is x= 1.75 then the equation can be written as $y= a(x- 1.75)^2+ c= ax^2+ 3.5ax+ 3.0625+ c= ax^2+ bx+ 22. We must have b= 3.5a.

If, in addition, y= 0 when x= -2, 0= 4a- 2b+ 22 so b= 11+ 2a.
 

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