# find the requested values. a) f(x)=x^2+4x+1standard form, vertex, x-intercept, y-intercept, domain, range,

### 4 Answers | Add Yours

Given `f(x)=x^2+4x+1`

(1) **In standard form you have **` x^2+4x+1=0 `

(2) One way to find the vertex is to identify the axis of symmetry. If the quadratic is in standard form, the axis is `x=(-b)/(2a)` , so the axis for this problem is `x=(-4)/2=>x=-2` .

The vertex lies on the axis of symmetry, so `f(-2)=(-2)^2+4(-2)+1=-3` and **the vertex is at (-2,-3).**

(3) The y-intercept is found by setting x=0.

(4) **The domain is all real numbers.** (This is true for all polynomials)((The domain is all allowable inputs))

(5) The range is all possible outputs. The graph of this quadratic is a parabola opening up, with a minimum at (-2,-3). **Thus the range is y>-3.**

`f(x)=x^2+4x+1`

First you need to find the axis of symmetry which will always be the x value of the vertex

in order to find the x value use the formula `-b/(2a)`

you can set out the numbers to make things easier `a=1` `b=4` `c=1`

`-4/(2(1)) ` `-4/2 ` ` x=-2`

plug in the x to get the y

`f(x)=-2^2+4(-2)+1`

`y=4-8+1`

`4-8=-4+1=-3 ` `y=-3`

the vertex is `(-2,-3)`

the aos is `-2`

domains is all real numbers

`y-in= -1`

range = `y>=-3` because a is positive

(-2,-3)

express x^2+4x+1 in the form of (x+a)^2+b

=(x+2)^2+1-(2)^2

=(x+2)^2-3

X=-2 and minimum point= -3

therefore curve start at (-2,3)

at x=0

y= 1 therefore the curve will intersect at y=1