Question

Hi,

Could somebody please explain and show their working for this question:
Sketch the graphs of these quadratic functions, in each case stating the vertex and the equation of the line of symmetry
1) y = 2(x +3)²

Woud I need to multiply out the brackets first? I'm not really sure where to start...

Answer 1

I'm assuming you've done parabolas before?

Answer 2

Yeah, kind of. I am working through quadratic functions at the moment and so have came across parabola's and the upside down U (whatever they may be called).
I am just not sure what I do with the 2 - do I need to multiply out the brackets first? As the questions says 'sketch' I assumed I didn't need to work it out to plot accurately but have looked at this type of question for expressions such as y=x√ - 5, where vertex would be (0,5) and symmetry the y axis...
Thank you

Answer 3

well, sorry for my lag, just I'm a bit lagged

Answer 4

well, the 2 needs no concern for the symmetry and the vertex, just to plot the points
usually that multiplier will simply "squeeze" the function graph

notice that the equation is already in "vertex form"
http://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms.png

so you can pretty much get the vertex from there

Answer 5

\(\bf y = 2(x +3)^2 \implies y = 2(x +3)^2+0\)

Answer 6

or if you wish \(\bf y = 2(x -(-3))^2 + 0\)

Answer 7

the multiplier in front of the binomial, that is the "2", is positive, that means the parabola opens upwards

the symmetry will occur at the vertex point, say