Parabola (1 Viewer)

[nrlwinner]

Hi. I'm just wondering how do you graph a formula (with axis and vertex) with the full equation of y = ax^2 + bx + c.

I've got a question here.

y = -x^2 + x +2

If someone could graph it and tell me how they did it step by step, I'd really appreciate it.

 

[Timothy.Siu]

Hi. I'm just wondering how do you graph a formula (with axis and vertex) with the full equation of y = ax^2 + bx + c.

I've got a question here.

y = -x^2 + x +2

If someone could graph it and tell me how they did it step by step, I'd really appreciate it.

u can try factorise it first

y=(-x+2)(x+1)
so now u can find the intercepts,
when y=0 x=2 or -1
when x=0 y=2

from this you can draw the basic shape of the parabola, noting that it is concave down because of the negative infront of the x^2
in yr 10 u learn that the axis is like x=-b/2a=1/2
alternatively u can differentiate it and find the turning point which is probably the preferred way to find the axis and vertex
y'=-2x+1
=0 when x=1/2, then y=9/4

 

[MC Squidge]

y = -x^2 + x +2
let x=0 to find y intercepts
y=2
therefore y intercept is (0,2)
let y=0 to find x intercepts
0 = -x^2 + x +2
x^2-x-2=0
(x-2)(x+1)=0
x=2,-1
therefore x intercept is (2,0) and (-1,0)
since it is y = -x^2 + x +2
it is a concave down parabola since the coefficient of x^2 is a negative number (in this case -1)
now to find the turning point, the final step
y = -x^2 + x +2
y'=-2x+1
y'=0 to find turning points
2x=1
x=(1/2)
f(1/2)=-((1/2)^2)+1/2+2=9/4
therefore maximum turning point @ (1/2,9/4)

putting all that info together

 

[nrlwinner]

Can you explain the vertex steps in more detail because I haven't done any work on it yet.

 

[MC Squidge]

to find the vertex of a parabola, u differentiate the function to find a maximum or a minimum. u solve this derivative equal to zero, letting y=0, solving for x. u then sub that x value into the original function to get the coordinates of the vertex

 

[jet]

The vertex occurs on the axis of symmetry. So, you just substitute the value for the axis of symmetry into the equation for the parabola and you have the x- and y-values for the vertex.
I am guessing you haven't done calculus yet, so the part that MC Squidge did wouldn't make sense.
Basically, for f(x) = ax^2 + bx + c, the axis of symmetry is the line x = -b/2a, and the vertex is the point (-b/2a, f(-b/2a))

 

[Drongoski]

Can you explain the vertex steps in more detail because I haven't done any work on it yet.

This is how I'd do it. (I carry around a rigid piece of clothe-hanger wire fashioned into a parabola):

Edit

Hell! I typed out the whole thing in LaTeX and lost it again!

My approach is: By completion of the square, we end up with:

y - 9/4 = -(x-1/2)²

This has the exact same graph as y = -x² except that the vertex is now at (1/2, 9/4) instead of (0, 0). I would physically move my rigid wire parabola (careful - don't tilt it) to its new address to demonstrate what happens.

 

[Drongoski]

Squidge . . . can u please explain to me child-like how to get a diagram up just like the one u put up. would be enormously useful to me. PM me perhaps. Thanks.