Question 194493
Since the equation has the x- intercepts (2,0) and (-7,0), this means that the roots are {{{x=2}}} or {{{x=-7}}}



{{{x=2}}} or {{{x=-7}}} Start with the given roots.



{{{x-2=0}}} or {{{x+7=0}}} Get everything to the left side.



{{{a(x-2)(x+7)=0}}} Combine the equations using the zero product property. Also, insert the variable "a" (this will be used to make the graph have a y-intercept of (0,5))



{{{a(x^2+5x-14)=0}}} FOIL




So the general equation with the x- intercepts (2,0) and (-7,0) is: {{{y=a(x^2+5x-14)}}}




{{{y=a(x^2+5x-14)}}} Start with the given equation.



{{{y=ax^2+5ax-14a}}} Distribute



{{{5=a*0^2+5a(0)-14a}}} Plug in {{{x=0}}} and {{{y=5}}} (from the point (0,5))



{{{5=-14a}}} Multiply and simplify



{{{5/(-14)=a}}} Divide both sides by -14.



{{{a=-5/14}}} Reduce and rearrange the equation.




{{{y=ax^2+5ax-14a}}} Go back to the previous equation



{{{y=-(5/14)x^2+5(-5/14)x-14(-5/14)}}} Plug in {{{a=-5/14}}}



{{{y=-(5/14)*x^2-(25/14)x+5}}} Multiply and simplify




So the quadratic equation is {{{y=-(5/14)*x^2-(25/14)x+5}}}



Here's a graph to visually confirm the answer:



{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-(5/14)*x^2-(25/14)x+5)

)}}}