‚Äči have to find the axis of symmetry and the vertex with a graph and show my work.

Mathematics
Tutor: None Selected Time limit: 1 Day

y=x^2-6x+11

Apr 29th, 2015

It is a parabola. It's vertex is its lowest point (minimum). The axis of symmetry is the vertical line passing the vertex.

Find minimum by differentiation: y'(x) = 2x - 6. This = 0 when x=3. y(3) = 9-18+11 = 2.

So: the vertex is (3; 2) and the axis of symmetry is x=3 (vertical line).

Please ask if something unclear.

Apr 29th, 2015

i don't really understand how you got that...

Apr 30th, 2015

It is a parabola because it has a form y = a*x^2 + bx + c. Probably you know that.
Such a parabola has a vertex at a minimum (if a>0 and branches go up) or minimum if a<0 and branches go down.

In both cases it is at x where y'(x) = 0 (probably also known). Then I differentiate and found the root.

There is another method: y(x) = x^2 - 6x + 11 = (x-3)^2 + 2. Because (x-3)^2 >= 0 and =0 only for x=3, then y(x) has a minimum of 2 at the point x=3. Also, if we change x to 6-x then y don't change, y(6-x) = y(x). So if a point (x; y) belongs to the parabola, then the point (6-x; y) also belongs. This is the same that "parabola is symmetric with respect to the line x=3".


Apr 30th, 2015

oooh okay thank you so much cxx

Apr 30th, 2015

please ask more if not clear

Apr 30th, 2015

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