Respuesta :
We know a quadratic function in standard form is given by:
[tex]y=ax^2+bx+c[/tex]where a, b and c are constants. To find the value of the constants we can use the points given.
For the point (-2,65) we have:
[tex]\begin{gathered} (-2)^2a-2b+c=65 \\ 4a-2b+c=65 \end{gathered}[/tex]For the point (0,165) we have:
[tex]\begin{gathered} (0)^2a+(0)b+c=165 \\ c=165 \end{gathered}[/tex]For the point (6,225) we have:
[tex]\begin{gathered} (6)^2a+6b+c=225 \\ 36a+6b+c=225 \end{gathered}[/tex]Hence we have the system of equations:
[tex]\begin{gathered} 4a-2b+c=65 \\ 36a+6b+c=225 \\ c=165 \end{gathered}[/tex]Plugging the value of c in the first two equations we have:
[tex]\begin{gathered} 4a-2b+165=65 \\ 36a+6b+165=225 \end{gathered}[/tex]which leads to:
[tex]\begin{gathered} 4a-2b=-100 \\ 36a+6b=60 \end{gathered}[/tex]Multiplying the first equation by 3 we have:
[tex]\begin{gathered} 12a-6b=-300 \\ 36a+6b=60 \end{gathered}[/tex]adding the equations we have that:
[tex]\begin{gathered} 48a=-240 \\ a=-\frac{240}{48} \\ a=-5 \end{gathered}[/tex]Plugging the value of a in the equation above we have:
[tex]\begin{gathered} 4(-5)-2b=-100 \\ -20-2b=-100 \\ 2b=100-20 \\ 2b=80 \\ b=\frac{80}{2} \\ b=40 \end{gathered}[/tex]Once we know the values of the constants we conclude that the quadratic equation in standard form is:
[tex]y=-5x^2+40x+165[/tex]Now, to write the equation in vertex form we need to complete the square on x:
[tex]\begin{gathered} y=-5(x^2-8x)+165 \\ y=-5(x^2-8x+(-\frac{8}{2})^2)+165+5(\frac{8}{2})^2 \\ y=-5(x-4)^2+245 \end{gathered}[/tex]Therefore, the equation written in vertex form is:
[tex]y=-5(x-4)^2+245[/tex]Now, the unique features of the two forms presented are:
Standard form: The equation is expanded and reduced.
Vertex form: The vertex of the parabola is readily shown in the equation.
