What is the equation of the parabola that has a vertex at # (3, -5) # and passes through point # (13,43) #?

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Answer 1 · 2016-08-29T20:07:29

#color(blue)("I have taken you to a point from which you can take over")#

Let the point #P_1->(x,y)=(13,43)#

Quadratic standard form equation: #y=ax^2+bx+5color(white)(" ").............................Eqn(1)#

Vertex form equation: #y=a(x+b/(2a))^2+kcolor(white)(" ") .......................Eqn(2)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Using Eqn(2)")#

We are given that Vertex#->(x_("vertex"),y_("vertex"))=(3,-5)#

But #x_("vertex")=(-1)xxb/(2a)=+3" "=>" "b=-6acolor(white)(" ")......Eqn(3)#

Side note: #k=-5# from vertex y-coordinate
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Using Eqn(3) substitute for b in Eqn(1)")#

#y=ax^2+(-6a)x+5# ...........................Eqn(4)

But we are given the point #P_1->(13,43)#

Thus Eqn(4) becomes:

#43=a(13)^2-6a(13)+5color(white)(" ")......Eqn(4_a)#

#color(blue)("From this you can solve for "a" and from that solve for "b)#

#color(red)("I will let you take over from this point")#