For (a); I tried this:
y = x^2 + qx + r
y = o^2 + 6 * 0 + 16
y = -16
y = x^2 -6x - 16
y = 8^2 - 6*8 - 16
y = 0
I am lost with (b) and not sure about (c)
I think we can get the minimum point by using Calculus.
So point P's coordinates is: (3 ; - 25)
Now this one really isn't hard.
Look at this:
Would you be able to solve this?
Of course you would!
But look at what i did:
I simply treated as
So let's solve this one:
But remember that is actually equal to
I don't think GAdams has taken a calculus class as yet. We can use the vertex formula to get the lowest Point.
This is an upward opening parabola, so the vertex occurs at the minimum point.
So the coordinates for the lowest point is