I need help to write an equation of a quadratic function 
whose graph is a parabola that has a vertex (-3,7) and 
that passes through the origin. Please.

The standard form of a parabola with vertex (h,k) is

y = a(x - h)² + k

where (h, k) is its vertex.

Since its vertex is (-3,7) we substitute -3 for h and 7 fo k.

y = a(x - (-3) )² + 7

y = a(x + 3)² + 7

Since it passes through the origin, which is the point
(x,y) = (0,0), if we substitute that point into the equation,
the equation must be satisfied, So we substitute that point 
by substituting 0 for x and 0 fot y:

y = a(x + 3)² + 7

0 = a(0 + 3)² + 7

0 = a(3)² + 7

0 = a(9) + 7

0 = 9a + 7

add -9a to both sides

-9a = 7

divide both sides by -9

  a = -7/9

So now we go back to 

y = a(x + 3)² + 7

and replace a by (-7/9)

y = (-7/9)(x + 3)² + 7

Your teacher may accept it that way, or you
can clear of fractions  by multiplying 
through by the LCD = 9

9y = -7(x + 3)² + 63

9y = -7(x + 3)(x + 3) + 63

9y = -7(x² + 3x + 3x + 9) + 63

9y = -7(x² + 6x + 9) + 63

9y = -7x² - 42x - 63 + 63

9y = -7x² - 42x, which you could write as 7x² + 42x + 9y = 0

or if you didn't want to do that you could

Factor out -7x on the right

9y = -7x(x + 6)

Divide through by 9

 y = -7(x + 6)/9

There are lots of ways you could leave the answer.

Edwin