Answer:
(5,0) or (1,0)
Step-by-step explanation:
1.) First you distribute the parenthesis (as it goes first in the PEMDAS series)
[tex]2(x-3)^{2} -8=2(x-3)(x-3)-8= 2(x^{2} -6x+9)-8[/tex]
2.) Distribute the 2 into the expression in the parenthesis and simplify
[tex]2(x^{2} -6x+9)-8=2x^{2} -12x+18-8=\\2x^{2} -6x+10=g(x)[/tex]
3.) Factor the quadratic
(there is many ways for different equations, so it is hard to explain how to, maybe go online if you are struggling, that's how I do mine)
[tex]2x^{2} -12x+10=2(x-5)(x-1)=0[/tex]
(Editors note: reason why we must put zero to solve for the functions zero is because we must find the "X" when "Y=0", or in this case "g(x)=0")
4.) Separate each of the parenthesis that you factored from the equation into two equations and solve for x when g(x)=0
[tex]2(x-5)=0\\2x-10=0\\2x=10\\\frac{2x}{(2)} =\frac{10}{(2)} \\x=5[/tex]
[tex](x-1)=0\\x=1[/tex]
So when y=0, x=5 or 1
5.) Turn the x and y values given into point form
(x=5,1) (y=0) so... (5,0) and (1,0)
