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Find equation of a quadratic with line of symmetry

doc5555

New member
Feb 20, 2019
1
mathhelp.jpg
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,120
Here's something to start you off. The axis of symmetry for the parabola \(\displaystyle y = ax^2 + bx + c\) is \(\displaystyle x = -\dfrac{b}{2a}\) and if an x-intercept is at (-2, 0) then \(\displaystyle 0 = a(-2)^2 + b(-2) + c\). That will give you both b and c as functions of a.

Can you finish?

-Dan
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
368
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.