Thread: y = f(x) = (x-5)²

y = f(x) = (x-5)²

Write an equation which represents x = f(y) and solve for y.

How to solve this? Thanks.

Originally Posted by shenton

y = f(x) = (x-5)²

Write an equation which represents x = f(y) and solve for y.

How to solve this? Thanks.

Hello,

if I understand your question correctly you should do:



EB

Thanks, I don't understand the question myself. I copied direct from the paper.

The topic heading is "Comparing the Graphs of y=f(x) and y=f-1(x) or x=f(y)"

y=f-1(x) is the inverse of y=f(x).

I don't know if x=f(y) is the same as y=f-1(x) or if it is a different thing.

Your answer makes sense. We know y = (x-5)² produce a "U" graph that was moved 5 units to the right.

This answer, f-1(x) = ±√(x) + 5, should produced a "U" shape rotated at 180 degrees but since the graph of radical functions has x-values > 0, I do not see the "U' shape.

Just some comments, sorry, I don't know what I'm taliking since I am not an expert in inverse functions of a quadratic expression.

Originally Posted by shenton

Thanks, I don't understand the question myself. I copied direct from the paper.

The topic heading is "Comparing the Graphs of y=f(x) and y=f-1(x) or x=f(y)"

y=f-1(x) is the inverse of y=f(x).

I don't know if x=f(y) is the same as y=f-1(x) or if it is a different thing.

Your answer makes sense. We know y = (x-5)² produce a "U" graph that was moved 5 units to the right.

This answer, f-1(x) = ±√(x) + 5, should produced a "U" shape rotated at 180 degrees but since the graph of radical functions has x-values > 0, I do not see the "U' shape.

...

Hello,

you wrote ...should produced a "U" shape rotated at 180 degrees.... Not quite, because then you get a parabola opened downward but not the graph of the inverse function. I attached a diagram with the original function and both inverse function of parts of f.
Similar colours mean that the corresponding graphs are reflected at the line y = x (and that's by the way the method you can "construct" point for point the inversed graph) To control your drawings (or sketches): If the graph of a function intersects with the line y = x then the graph of the inverse function must do so too.

EB

You are right about the graph of the inverted function.

If the graph of a function intersects with the line y = x then the graph of the inverse function must do so too - Something new for me, that is good to know.

We worked out the inverse of y = (x-5)² is y-1 = ±√(x) + 5. I know the graph is correct since the vertex of the function is (0,5) and the vertex of the inverse function is (5,0).

What is strange is that √(x) + 5 will not produce the "U" shape inverted graph. Algebratically, it looks right. Graphically, it looks right. I am stumped now.

Originally Posted by shenton

...

What is strange is that √(x) + 5 will not produce the "U" shape inverted graph. Algebratically, it looks right. Graphically, it looks right. I am stumped now.

Hello,

it's me again.

The U you are looking for opens to the right and is composed of 2 parts:
y = √(x) + 5 is the upper part because you add a value to 5

y = -√(x) + 5 is the lower part because you subtract a value from 5.

You can easily detect the u-shaped curve in my diagram.

EB