find the requested values. a) f(x)=x^2+4x+1standard form, vertex, x-intercept, y-intercept, domain, range,
(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.
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
`4-8=-4+1=-3 ` `y=-3`
the vertex is `(-2,-3)`
the aos is `-2`
domains is all real numbers
range = `y>=-3` because a is positive
express x^2+4x+1 in the form of (x+a)^2+b
X=-2 and minimum point= -3
therefore curve start at (-2,3)
y= 1 therefore the curve will intersect at y=1