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...
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.**
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