SOLUTION: Find the quadratic function y=f(x) whose graph has a vertex (-3,2) and passes through the point (-1,0). Write the function in standard form.

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Question 1176621: Find the quadratic function y=f(x) whose graph has a vertex (-3,2) and passes through the point (-1,0). Write the function in standard form.
Found 3 solutions by ewatrrr, MathLover1, ikleyn:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!

Hi Disregard comments, both tutors with same solution. :) Using the vertex form of a parabola, where(h,k) is the vertex vertex (-3,2) y = a(x+3)^2 + 2 P(-1,0) 0 = 4a+ 2 -2/4 = a = -.5 and Standard form: y = -.5(x^2 + 6x + 9) + 2 = -.5x^2 + 3x -9/2 + 2 = -.5x^2 + 3x -5/2 Wish You the Best in your Studies.

Answer by MathLover1(20849)   (Show Source):
You can put this solution on YOUR website!
the quadratic function in vertex form, some refer to it as “standard form”
, where (, ) is the vertex of the parabola
if given a vertex (,) then and

if passes through the point (,) we have

and your equation is:

Answer by ikleyn(52750)   (Show Source):
You can put this solution on YOUR website!
.

You have two solutions, one from @MathLover1 and another from @ewatrrr.

The solution from @MathLover1 is correct.

The answer from @ewatrrr is INCORRECT.

SOLUTION: Find the quadratic function y​=f(x) whose graph has a vertex ​(​-3,​2) and passes through the point ​(-1​,0). Write the function in standard form.

SOLUTION: Find the quadratic function y=f(x) whose graph has a vertex (-3,2) and passes through the point (-1,0). Write the function in standard form.