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The sketch below shows a graph with the equation y = ab", where b > 0.'

Work out the value of k.
(-1,k)
(2,28)
(6, 448) *

The sketch below shows a graph with the equation y = ab", where b > 0.' Work-example-1

1 Answer

6 votes

The k = 0 or k = 14. Since the graph passes through the point (-1, k), and k cannot be negative, we can conclude that k = 0 .

Since the graph of y = ab^x passes through the point (6, 448), we can write the equation:

448 = ab^6

Similarly, since the graph also passes through the point (-1, k), we can write the equation:

k = ab^-1

Dividing the first equation by the second equation, we get:

448 / k = ab^6 / ab^-1

Simplifying, we get:

448 / k = b^7

Since b > 0, we can take the square root of both sides to get:

b = sqrt(448 / k)

Substituting this value of b into the equation y = ab^x, we get:

y = ab^x = sqrt(448 / k) * ab^x

Since the graph passes through the point (2, 28), we can substitute x = 2 and y = 28 into this equation to get:

28 = sqrt(448 / k) * ab^2

Dividing both sides by sqrt(448 / k), we get:

28 / sqrt(448 / k) = ab^2

Simplifying, we get:

28 sqrt(k / 448) = ab^2

Since ab^2 is a positive number, we can take the square root of both sides to get:

sqrt(28 sqrt(k / 448)) = ab

Simplifying, we get:

sqrt(14k) = ab

Substituting this value of ab into the equation k = ab^-1, we get:

k = sqrt(14k) * -1

Squaring both sides, we get:

k^2 = 14k

Subtracting 14k from both sides, we get:

k^2 - 14k = 0

Factoring, we get:

k(k - 14) = 0

Therefore, k = 0 or k = 14. Since the graph passes through the point (-1, k), and k cannot be negative, we can conclude that k = 0.

User MichaelH
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