Locus & the Parabola: finding eq. of a parabola (1 Viewer)

[Makro]

Find the equation of the following:

c) the line x=2 as axis, vertex (2,4) and crosses the y axis at y=8.

Not sure what the y-intercept has to do with it, can someone explain?

 

[Iruka]

You need to know how fast the parabola is increasing (and whether it is concave up or concave down) in order to uniquely specify the equation.

For example, the parabola y-4=-(x-2)^2 fullfils two of those conditions but not the last one.

 

[AlexJB]

Equation of a parabola is (x-h)^2 = 4a(y-q)
Vertex is (2,4) therefore we know the formula is: (x-2)^2 = 4a(y-4)
All we need is the value for a and we have the equation?

This was year 11 wasn't it? I forget how to find the value of a. I know I could do it using algebra and just manipulating the above equation, but I'm sure theres another way.

Eh:
When x=0 y=8.
(0-2)^2 = 4a(8-4)
4 = 4a*4
16a = 4
a = 1/4

Therefore equation is: (x-2)^2 = y-4

Edit: addressing the above poster. Since we're given the vertex and the x-intercept we can see its concave up.

 

[Makro]

Thanks for the help. Yes it's year 11 work, but my school is doing that now. We didn't fit in Quadratic Function + Locus in 3 terms, so we started with them, just finished today and soon starting probability.

Thanks again.