Given ` x^2+5x+7 ` :

(i) Find the vertex:

(a) Rewrite in vertex form:

`x^2+5x+7 `

Add and subtract the square of 1/2 the linear term, `(5/2)^2`, to get ` =x^2+5x+25/4-25/4+7 `

`=(x+5/2)^2+3/4 `

is now in vertex form with vertex `(-5/2,3/4) `

(b) Or find the axis of symmetry ` x=(-b)/(2a) ` so `x=(-5)/(2) ` ; The value of the expression at this point is 3/4 so the vertex is at `(-5/2,3/4) `

(ii) The line of symmetry is given by `x=(-b)/(2a) ` where the expression is given as `ax^2+bx+c ` ; here a=1, b=5, and c=7. So the axis (line) of symmetry is ` x=-5/2` .

(iii) The graph is a parabola, opening up, normal width, with vertex `(-5/2,3/4) `

The y-intercept can be found by setting x=0 to get 7. Some points (found by choosing convenient x-values near the vertex and symmetry) include (0,7),(-1,3),(-2,1),(-3,1), (-4,3),(-5,7)

The graph:

** The vertex form is `y=(x+5/2)^2+3/4 ` . The 5/2 shifts the graph of `y=x^2` left 2.5 units; the `3/4` term shifts the graph of `y=x^2` up`3/4` of a unit. Since the leading coefficient is 1 there is no vertical stretch/compression.**

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.

**Further Reading**