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.

Algebra ->  Quadratic Equations and Parabolas -> 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.      Log On


   

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) About Me  (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, y=a%28x-h%29%5E2+%2Bk 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

y+=+-.5%28x%2B3%29%5E2+%2B+2
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(20850) About Me  (Show Source):
You can put this solution on YOUR website!
the quadratic function in vertex form, some refer to it as “standard form”
f+%28x%29+=+a%28x+-+h%29%5E2+%2B+k, where (h, k) is the vertex of the parabola
if given a vertex ​(​-3,​2) then h=-3 andk=2
f+%28x%29+=+a%28x+-+%28-3%29%29%5E2+%2B+2
f+%28x%29+=+a%28x+%2B3%29%5E2+%2B+2
if passes through the point ​(-1​,0) we have
0+=+a%28-1+%2B3%29%5E2+%2B+2
-2+=+a%282%29%5E2+
-2=+4a
a=-2%2F4
a=-1%2F2
and your equation is:
f+%28x%29+=+-%281%2F2%29%28x+%2B3%29%5E2+%2B+2







Answer by ikleyn(52903) About Me  (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.