Answer:
Equation for the quadratic in vertex form: [tex]y-2=2\,(x-1)^2[/tex]
Equation in standard form: [tex]y=2x^2-4x+4[/tex]
Step-by-step explanation:
Recall that there is a form for writing the general equation of a parabola in vertex form:
[tex]y-y_{vertex}=a(x-x_{vertex})^2[/tex]
so in our case, considering the the coordinates of the vertex (1,2) are given, we have:
[tex]y-y_{vertex}=a(x-x_{vertex})^2\\y-2=a(x-1)^2[/tex]
Now, we can find the parameter "a" missing for the general equation, by using the information on a point (3,10) through which the parabola passes:
[tex]y-2=a\,(x-1)^2\\(10)-2=a],(3-1)^2\\8=a\, (2)^2\\8=4\,a\\a=2[/tex]
So now we have the general form of this quadratic:
[tex]y-2=2\,(x-1)^2[/tex]
which solving for "y" in order to give its standard form results in:
[tex]y=2(x-1)^2+2\\y=2\,(x-1)^2+2\\y=2\,(x^2-2x+1)+2\\y=2x^2-4x+2+2\\y=2x^2-4x+4[/tex]
