Answer:
[tex]\large\boxed{y=4(x-3)^2+2\ \bold{vertex\ form}}\\\boxed{y=4x^2-24x+38\ \bold{standard\ form}}[/tex]
Step-by-step explanation:
The vertex form of a parabola:
[tex]y=a(x-h)^2+k[/tex]
(h, k) - vertex
We have the vertex at (3, 2) → h = 3 and k = 2.
Substitute:
[tex]y=a(x-3)^2+2[/tex]
The point (4, 6) is on athe parabola. Put the coordinates of this point to the equation:
[tex]6=a(4-3)^2+2[/tex] subtract 2 from both sides
[tex]6-2=a(1)^2+2-2[/tex]
[tex]4=a\to a=4[/tex]
Finally:
[tex]y=4(x-3)^2+2[/tex] vertex form
use (a - b)² = a² - 2ab + b²
[tex]y=4(x^2-6x+9)+2[/tex] use the distributive property
[tex]y=4x^2-24x+36+2[/tex]
[tex]y=4x^2-24x+38[/tex] standard form
Answer:
y=4(x-3)^2+2
Step-by-step explanation:
Hopefully this helps :)
