What is the equation of the parabola that has a vertex at # (-1, 16) # and passes through point # (3,20) #?

The standard form of the equation of a parabola is:
#f(x) = a(x-h)^2 + k#

From the question we know two things.

1. The parabola has a vertex at #(-1, 16)#
2. The parabola passes through the point #(3, 20)#

With those two pieces of information, we can construct our equation for the parabola.

Let's start off with the basic equation:
#f(x) = a(x-h)^2 + k#

Now we can substitute our vertex coordinates for #h# and #k#

The #x# value of your vertex is #h# and the #y# value of your vertex is #k#:

#f(x) = a(x+1)^2 + 16#

Note that putting #-1# in for #h# makes it #(x-(-1))# which is the same as #(x+1)#

Now substitute the point the parabola passes through for #x# and #y# (or #f(x)#):

#20 = a(x+1)^2 + 16#

Looks good. Now we have to find #a#

Combine all like terms:

Add 3 + 1 inside the parentheses:
#20 = a(4)^2 + 16#

Square 4:
#20 = 16a+ 16#

Factor out 16:
#20 = 16(a+1)#

Divide both sides by 16:
#20/16 = a+1#

Simplify #20/16#:
#5/4 = a+1#

Subtract 1 from both sides:
#5/4 -1 = a#

The LCD of 4 and 1 is 4 so #1 = 4/4#:
#5/4 -4/4 = a#

Subtract:
#1/4 = a#

Switch sides if you want:
#a = 1/4#

Now that you've found #a#, you can plug it into the equation with the vertex coordinates:

#f(x) = 1/4(x+1)^2 + 16#

And that's your equation.
Hope this helped.

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#

#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#

#"here "(h,k)=(-1,16)#

#rArry=a(x+1)^2+16#

#"to find a substitute "(3,20)" into the equation"#

#20=16a+16rArra=1/4#

#rArry=1/4(x+1)^2+16larrcolor(red)"in vertex form"#