How do you write a quadratic equation with vertex of (3,-6) and the point (-1,10) ?

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Answer 1 · 2017-03-27T15:29:31

I have taken you to a point where you can finish it off.

Consider the standardised form of #y=ax^2+bx+c#

Then #x_("vertex")=(-1/2)xxb#

#=># given that vertex#->(x,y)=(3-6)# then we have

#x_("vertex")=(-1/2)xxb=3#

#=>color(magenta)(b=3xx(-2/1)=-6)#

So we now have:

#y=ax^2-6x+c#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
2 unknowns and 2 sets of values so solvable.

For: #(x,y)->(3,-6)#
#=> -6=a(3^2)-6(3)+c" "...Equation(1)#

For: #(x,y)->(-1,10)#
#=>10=a(-1)^2-6(-1)+c.Equation(2)#

#-6=9a-18+c" "...Equation(1)#
#color(white)(.)10=color(white)(.)a+color(white)(.)6+c" "..Equation(2)#

Multiply #Equation(2)# by 9

#color(white)(.)90=9a+54+9c" "..Equation(2_a)#
#ul(-6=9a-18+c" ".....Equation(1))larr" subtract"#
#84=0a+72+10c#

#10c=12#

#color(magenta)(c=6/5)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now that you have #c# yo can determine the value of #a# by substitution. I will let you do that.