what are the coordinates of the vertex of the parabola defined by:
f(x)=9(x-8)+8
I'm confused on how to solve this. Help pls
Need an x^2 somewhere
perhaps
y = 9 (x-8)^2 + 8
or something?
If it is indeed
y = 9 (x-8)^2 + 8
then
y = 9 (x^2 - 16 x + 64) + 8
y = 9 x^2 - 144 x + 576 + 8
y = 9 x^2 - 144 x + 584
then
x^2 - 16 x + 64 8/9 = y/9
x^2 - 16 x = y/9 -64 -8/9
x^2 - 16 x + 64 = y/9 -64 -8/9 + 64
(x-8)^2 = y/9 -8/9 = (1/9)(y-8)
vertex at (8,8)
of course we could have short cut that by saying
9 (x-8)^2 = y-8
then
(x-8)^2 = (1/9)(y-8)
but best do the whole completing the square process.
To find the coordinates of the vertex of a parabola in the form f(x) = ax^2 + bx + c, you can use the formula:
x = -b / (2a)
In this case, the equation f(x) = 9(x-8) + 8 is already in the form f(x) = ax + b, so a = 9 and b = -72.
Now, let's substitute these values into the formula to find x:
x = -(-72) / (2 * 9)
= 72 / 18
= 4
So, the x-coordinate of the vertex is 4.
To find the y-coordinate of the vertex, substitute this x-value back into the original equation:
f(x) = 9(x-8) + 8
f(4) = 9(4-8) + 8
f(4) = 9(-4) + 8
f(4) = -36 + 8
f(4) = -28
Therefore, the coordinates of the vertex are (4, -28).