Fibd dilation factor of a transformed parabola, when vertex 3, 30 and x intercepts 0,0 and 6,0
2 answers
Give answer in y=a(x-h)^2 +k
Using the intercepts, you have
y = a(x-0)(x-6) = a(x^2 - 6x)
Since the vertex is midway between the roots, you know that it is at x=3
Since the vertex of x^2-6x is at (3,-9) and you want it to be at (3,30), you can see that a = -30/9 = -10/3
so now you need to put that into vertex form
x^2-6x+9 = (x-3)^2
So, now you need
y = -10/3 (x^2-6x+9-9) = -10/3 (x-3)^2 + 30
You could, of course, come at it in other ways, but this is the way I handled it.
y = a(x-0)(x-6) = a(x^2 - 6x)
Since the vertex is midway between the roots, you know that it is at x=3
Since the vertex of x^2-6x is at (3,-9) and you want it to be at (3,30), you can see that a = -30/9 = -10/3
so now you need to put that into vertex form
x^2-6x+9 = (x-3)^2
So, now you need
y = -10/3 (x^2-6x+9-9) = -10/3 (x-3)^2 + 30
You could, of course, come at it in other ways, but this is the way I handled it.
1 answer