Question 187247
"If the zeroes of a quadratic function are -4 and 6" means that the solutions to the equation {{{ax^2+bx+c=0}}} are {{{x=-4}}} or {{{x=6}}} (ie if you plug in {{{x=-4}}} or {{{x=6}}} into {{{ax^2+bx+c}}}, you will get zero)





{{{x=-4}}} or {{{x=6}}} Start with the given solutions



{{{x+4=0}}} or {{{x-6=0}}} Get the terms to the left side (so the right side is zero)



{{{(x+4)(x-6)=0}}} Combine the equations using the zero product property



Note: the zero product property states that if {{{A*B=0}}} then either {{{A=0}}} or {{{B=0}}} (or possibly both)



{{{a(x+4)(x-6)=0}}} Place an additional variable "a" onto the front of the equation. Note: this will be needed to satisfy the third condition.



{{{a(x^2-6x+4x-24)=0}}} FOIL



{{{a(x^2-2x-24)=0}}} Combine like terms.



So the equation that has the zeros of -4 and 6 is {{{y=a(x^2-2x-24)}}}



Now there's an additional condition: The y-intercept is 12



This means that when {{{x=0}}}, {{{y=12}}}



{{{y=a(x^2-2x-24)}}} Start with the previous equation.



{{{12=a(0^2-2(0)-24)}}} Plug in {{{x=0}}} and {{{y=12}}}



{{{12=a(0-2(0)-24)}}} Square 0 to get 0



{{{12=a(0-0-24)}}} Multiply



{{{12=a(-24)}}} Combine like terms.



{{{12/(-24)=a}}} Divide both sides by -24 to isolate "a".



{{{-1/2=a}}} Reduce



So the value of "a" is {{{a=-1/2}}}



This means that the equation then goes from {{{y=a(x^2-2x-24)}}} to {{{y=(-1/2)(x^2-2x-24)}}}



Now distribute to get: {{{y=(-1/2)(x^2)-(-1/2)2x-(-1/2)24}}}



and multiply: {{{y=(-1/2)x^2+x+12}}}



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Answer:



So the equation with the zeros -4 and 6 with the y-intercept of 12 is:



{{{y=(-1/2)x^2+x+12}}}



Here's the graph to visually verify our answer:



{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -5, 15,(-1/2)x^2+x+12)

)}}} 


Graph of {{{y=(-1/2)x^2+x+12}}} with x-intercepts (-4,0) and (6,0) with the y-intercept (0,12)