How do you find the equation for a parabola if you have (3,0), (5,0) and (0, 15)?

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Answer 1 · 2016-06-14T06:16:55

#y=x^2-8x+15#

Standard form of equation of a parabola is #y=ax^2+bx+c#

As it passes through points #(3,0)#, #(5.0)# and #(0,15)#, each of these points satisfies the equation of parabola and hence

#0=a*9+b*3+c# or #9a+3b+c=0# ........(A)
#0=a*25+b*5+c# or #25a+5b+c=0# ........(B)
and #15=a*0+b*0+c# or #c=15# ........(C)

Now putting (C) in (A) and (B), we get\

#9a+3b=-15# or #3a+b=-5# and .........(1)

#25a+5b=-15# or #5a+b=-3# .........(2)

Subtracting (1) from (2), we get #2a=2# or #a=1#

and hence #b=-5-3*1=-8#

Hence equation of parabola is

#y=x^2-8x+15# and it appears as shown below

graph{x^2-8x+15 [-5.5, 14.5, -2, 8.84]}