Question

give the vertex and x-intercepts of f(x) -x squared +3x +2

Answer 1

is this f(x) = x^2 +3x + 2?

Answer 2

yes

Answer 3

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.

Answer 4

or you can write the parabola in the form f(x)=a(x-h)^2+k

Answer 5

vertex=(h,k)

Answer 6

You need to know how to complete the square though

Answer 7

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)

Answer 8

oops but i did something wrong

Answer 9

I was about to say myin, but you are more intelligent than I. Fix it nao! :)

Answer 10

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

Answer 11

so the vertex is (-3/2,1/4) if I didn't screw up again

Answer 12

darn it (-3/2,-1/4)

Answer 13

and laxad found the x-intercepts above