How do I write transformation equations?

Question

The graph of G(x)=1/f(x), where f(x) is a quadratic function, is shown. The y-intercept of G(x) is 1/9 and the maximum point of G(x) is at (2,1).

The graph sorta looks like ahill which does NOT cross the x-axis.

Determine the equation of f(x) in general form.

Could you please explain show steps in solving the questions or at least explain how you came up with the answer.

Answer 1

G(x) = 1/f(x)

Since G doesn't cross the x-axis, neither will f. (I.e. f(x) will have no real roots. If they did...f would be 0 so G would be 1/0 or infinity.)

Since G is always positive, f will be too.

You're given the point (2, 1) as the maximum of G (so it will be the minimum of f).

Since G(2) = 1, f(2) = 1/G(1) = 1/1 = 1, so the vertex of f(x) will be (2, 1).

So now we've got (using the general form for a parabola):

f(x) = a(x - 2)² + 1

All that's left is to find a... and you find that from the y-intercept (0, 1/9).

G(0) = 1/9, so f(0) = 1/G(0) = 1/(1/9)= 9

So the point (0, 9) must fit the equation.

9 = a(0 - 2)² + 1
9 = 4a + 1
8 = 4a
2 = a

So

f(x) = 2(x - 2)² + 1 = 2(x² - 4x + 4) + 1 = 2x² - 8x + 9

Answer 2

when x=0, G(x)=1/9
Therefore, when x=0, f(x)=9

The maximum of G(x) is at (2,1)
Therefore, the minimum of f(x) is at (2,1)

So we have two points: (0,9) and a min at (2,1)

The min at (2,1) tells me I can write the function as

f(x) = a (x-2)^2 + 1

Moving over 2 from the min moves us up 8, so the quadratic coefficient is going to be 8 divided by 2 squared, or 8 over 4, or 2

f(x)=2(x-2)^2 + 1

f(x) = 2x^2 - 8x + 9