A curve has equation y= x^2 + 5x + 7 (i) Find the coordinates of the vertex of the curve  (ii) State the equation of the line of symmetry of the curve (iii) sketch the curve, stating the value of the intercept on the y-axis. Hi, I'm just very confused in regards of what I have to do with these questions and which method specifically to use?! Could you please show multiple methods if there are any please :)  and in regards to the graph you dont have to do it but could you please guide me on how to do it? thank you :) Also, if you have time, describing this question would REALLY help me: describe the geometrical transformation that maps the graph of y=x^2 on to the graph y= x^2 + 5x + 7. Thank you for all your help.       

Expert Answers

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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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