Question

Write an equation of a parabola in standard form that has a vertex of (-3,-4) and passes through (1,4)

Answer 1

As we know vertex -

$y=a(x+3)^2-4$

Passes through (1,4). Substitute x = 1 and y = 4

$4=a(1+3)^2-4\\4=a(4)^2-4\\4=16a-4\\4+4=16a\\16a=8\\a=\frac{8}{16}\\a=\frac{1}{2}$

Therefore the vertex equation is $y=\frac{1}{2}(x+3)^2-4$

Since you want the "standard" form then..

$y=\frac{1}{2}(x^2+6x+1)\\y=\frac{1}{2}x^2+3x+\frac{1}{2}$

So the standard form is 1/2x^2 + 3x + 1/2