anonymous
  • anonymous
give the vertex and x-intercepts of f(x) -x squared +3x +2
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
is this f(x) = x^2 +3x + 2?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok so to find the x intercepts we have to factor the quadratic. so if we factor the quadratic we will get the following: (x+2)(x+1) therefore x = -1, -2. so they are your x intercepts. now to find the vertex the function will be at minimum. so we will have ... to find the min or max of a function you have to take the derivative. so derivative of this quad is 2x + 3 = 0 so 2x = -3 and x = -3/2 Hence the vertex of this is (-3/2, -2) i believe.

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myininaya
  • myininaya
or you can write the parabola in the form f(x)=a(x-h)^2+k
myininaya
  • myininaya
vertex=(h,k)
myininaya
  • myininaya
You need to know how to complete the square though
myininaya
  • myininaya
so f(x)=x^2+3x+(3/2)^2+2+(3/2)^2=(x+3/2)^2+2+9/4=(x+3/2)^2+8/4+9/4=(x+3/2)^2+17/4 so the vertex is (-3/2,17/4)
myininaya
  • myininaya
oops but i did something wrong
anonymous
  • anonymous
I was about to say myin, but you are more intelligent than I. Fix it nao! :)
myininaya
  • myininaya
f(x)=x^2+3x+(3/2)^2+2-(3/2)^2=(x+3/2)^2+2-9/4=(x+3/2)^2+8/4-9/4=(x+3/2)^2-1/4
myininaya
  • myininaya
so the vertex is (-3/2,1/4) if I didn't screw up again
myininaya
  • myininaya
darn it (-3/2,-1/4)
myininaya
  • myininaya
and laxad found the x-intercepts above

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