Question

7.

x	y
-1	-50
0	-100
1	-200
2	-400

Explanation:

Step1: Check for exponential pattern

Let's assume the function is of the form \( y = ab^x + c \). When \( x = 0 \), \( y=- 100\), so \( a\times b^0 + c=-100\), which simplifies to \( a + c=-100 \).
When \( x = 1 \), \( y=-200\), so \( ab + c=-200 \).
When \( x = 2 \), \( y=-400\), so \( ab^2 + c=-400 \).

Subtract the equation for \( x = 0 \) from \( x = 1 \): \( ab + c-(a + c)=-200-(-100)\)
Simplify: \( ab - a=-100\), \( a(b - 1)=-100 \)

Subtract the equation for \( x = 1 \) from \( x = 2 \): \( ab^2 + c-(ab + c)=-400-(-200)\)
Simplify: \( ab^2 - ab=-200\), \( ab(b - 1)=-200 \)

Now divide the second new equation by the first: \( \frac{ab(b - 1)}{a(b - 1)}=\frac{-200}{-100}\) (assuming \( a
eq0 \) and \( b
eq1 \))
We get \( b = 2 \)

Substitute \( b = 2 \) into \( a(b - 1)=-100 \): \( a(2 - 1)=-100\), so \( a=-100 \)

Now from \( a + c=-100 \), substitute \( a=-100 \): \( - 100 + c=-100\), so \( c = 0 \)

So the function is \( y=-100\times2^x \)

Let's verify with \( x=-1 \): \( y=-100\times2^{-1}=-100\times\frac{1}{2}=-50 \), which matches the table.

Step2: Confirm the pattern

We can see that each time \( x \) increases by 1, \( y \) is multiplied by 2 (since \( b = 2 \)). When \( x = 0 \), \( y=-100 \), which is the initial value. The general form of an exponential function is \( y = ab^x \) (since \( c = 0 \)) with \( a=-100 \) and \( b = 2 \), so \( y=-100(2)^x \)

Answer:

The relationship between \( x \) and \( y \) is given by the exponential function \( y = - 100\times2^{x} \) (or \( y=-100(2)^x \))