SOLUTION: Find the standard form f(x)=a(x-h)^2+k of the quadratic function that has a vertex at (-1/4,-1/6) and x-intercept of -3/4

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Find the standard form f(x)=a(x-h)^2+k of the quadratic function that has a vertex at (-1/4,-1/6) and x-intercept of -3/4      Log On


   

Question 1087418: Find the standard form f(x)=a(x-h)^2+k of the quadratic function that has a vertex at (-1/4,-1/6) and x-intercept of -3/4
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
Find the standard form f%28x%29=a%28x-h%29%5E2%2Bk of the quadratic function that has a vertex at
(-1%2F4,-1%2F6) =>h=-1%2F4 and k=-1%2F6
so far, equation is:
f%28x%29=a%28x-%28-1%2F4%29%29%5E2%2B%28-1%2F6%29
f%28x%29=a%28x%2B1%2F4%29%5E2-1%2F6

and if x-intercept of -3%2F4+=> x=-3%2F4+ when f%28x%29=0+
use it to find a:
0=a%28-3%2F4%2B1%2F4%29%5E2-1%2F6
0=a%28-2%2F4%29%5E2-1%2F6
1%2F6=a%28-1%2F2%29%5E2
1%2F6=a%281%2F4%29
a=%281%2F6%29%2F%281%2F4%29
a=%284%2F6%29
a=%282%2F3%29
so, your equation is:
f%28x%29=%282%2F3%29%28x%2B1%2F4%29%5E2-1%2F6






Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Find the standard form f(x)=a(x-h)^2+k of the quadratic function that has a vertex at (-1/4,-1/6) and x-intercept of -3/4
f%28x%29+=+y+=+a%28x+-+h%29%5E2+%2B+k

0+=+a%28-+3%2F4+-+-+1%2F4%29%5E2+-+1%2F6 ------- Substituting 

0+=+a%28-+3%2F4+%2B+1%2F4%29%5E2+-+1%2F6

1%2F6+=+a%28-+1%2F2%29%5E2

1%2F6+=+%281%2F4%29a

a+=+%281%2F6%29%2F%281%2F4%29

matrix%281%2C5%2C+a%2C+%22=%22%2C+%281%2F6%29+%2A+4%2C+%22=%22%2C+2%2F3%29

highlight_green%28f%28x%29+=+%282%2F3%29%28x+%2B+1%2F4%29%5E2+-+1%2F6%29 ------- Substituting