I am having some trouble solving this exercise.
I have a family of fuctions defined by $y=k\cdot 2^{2px - 1}-3$
The function and graphic is $f(x)=2^{2x-2}-3$
And two dots of the ghaphic f: A (2;1) and B (1;-2)
I need to prove that $f(x)=2^{2x-2}-3$
Please, give me a hint.
Thank you :)
On
You just need to make equations with given data.
Point A has coordinates $x=2, y=1$, so if it belongs to the function's graph, the function equation is satistied by those coordinates: $$f(2) = k\cdot 2^{2p\cdot 2 - 1}-3 = 1$$ Similary for point B you have $x=1, y=-2$ satisfy $y=f(x)$, hence $$f(1) = k\cdot 2^{2p\cdot 1 - 1}-3 = -2$$
Add $3$ to both sides of both equations and you have:
$$k\cdot 2^{2p\cdot 2 - 1} = 4$$ $$k\cdot 2^{2p\cdot 1 - 1} = 1$$
Separate $(-1)$ from exponents: $$k\cdot \left(2^{2p\cdot 2} / 2 \right) = 4$$ $$k\cdot \left(2^{2p\cdot 1} / 2 \right) = 1$$
Now use the known relation of exponents: $(A^B)^C = A^{B C}$: $$k\cdot \left((2^p)^4 / 2 \right) = 4$$ $$k\cdot \left((2^p)^2 / 2 \right) = 1$$
Divide them side-wise: $$\frac {k\cdot \left((2^p)^4 / 2 \right)}{k\cdot \left((2^p)^2 / 2 \right)} = \frac 41$$
$$\frac {(2^p)^4}{(2^p)^2} = \frac 41$$ $$(2^p)^2 = 4 = 2^2$$ $$2^p = 2 = 2^1$$ $$p = 1$$
Now recall the equation for point B: $$k\cdot 2^{2p\cdot 1 - 1}-3 = -2$$
Plug $p=1$ to get: $$k\cdot 2^{2\cdot 1\cdot 1 - 1} = 1$$ $$k\cdot 2^{2 - 1} = 1$$ $$k\cdot 2^1 = 1$$ $$k = 1/2$$
Finally $$\begin{align} f(x) && = k\cdot 2^{2px - 1}-3 \\ && = \frac 12 \cdot 2^{2x - 1}-3 \\ && = 2^{2x - 1 - 1}-3 \\ && = 2^{2x - 2}-3 \\ \end{align}$$
Here's a nudge.
From your two points you get:
$$1 = k \cdot 2^{4p-1} - 3$$ $$-2 = k \cdot 2^{2p-1} - 3$$
Consolidate the constants:
$$4 = k \cdot 2^{4p-1}$$ $$1 = k \cdot 2^{2p-1}$$
This implies
$$k \cdot 2^{4p-1} = 4k \cdot 2^{2p-1}$$
Divide both sides by $k \cdot 2^{2p-1}$.
Can you take it from here?