Question 155148

Since there are two real solutions, this means that the discriminant is greater than zero. Since the x-intercepts are (-1,0) and (0.67,0) this means that the solutions are {{{x=-1}}} or {{{x=0.67}}}


{{{x=-1}}} or {{{x=0.67}}} Start with the given solutions



{{{x+1=0}}} or {{{x-0.67=0}}} Get every term to the left side



{{{a(x+1)(x-0.67)=0}}} Use the zero product property to join the two equations. I'm introducing the term "a" to make sure that the quadratic goes through the point (0,-2)



{{{a(x^2-0.67x+x-0.67)=0}}} FOIL



{{{a(x^2+0.33x-0.67)=0}}} Combine like terms.



So the equation becomes {{{y=a(x^2+0.33x-0.67)}}}



{{{-2=a(0^2+0.33(0)-0.67)}}} Plug in {{{x=0}}} and {{{y=-2}}}



{{{-2=a(-0.67)}}} Multiply and simplify. Notice how everything cancels out but the "0.67"



{{{-2=a(-0.67)}}} Multiply and simplify. Notice how everything cancels out but the "0.67"



{{{3=a}}} Divide both sides by -0.67 to isolate "a"


So the constant is {{{a=3}}}


{{{y=3(x^2+0.33x-0.67)}}} Plug in {{{a=3}}} into the equation



{{{y=3(x^2)+3(0.33x)-3(0.67)}}} Distribute



{{{y=3x^2+x-2}}} Multiply


So the equation that goes through the three points is {{{y=3x^2+x-2}}}