3 years ago

What is (7,3) and move 5 units left.

Geography

1 answer:

marshall27 [118]3 years ago

8 0

I believe it's going to be (2,3)

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On a coordinate plane, 2 exponential functions are shown. f (x) = 5 Superscript x Baseline approaches y = 0 in quadrant 2 and in

kiruha [24]

Answer:

$k = -7$

Explanation:

Given

$f(x) = 5^x$ through (0,1) and (1,5)

$g(x) = 5^x + k$ through (0,-6) and (1,-2)

Required

Determine the value of k if f(x) is translated vertically down

To get the value of k, we perform the following operations.

For (0,-6) --- Substitute 0 for x and -6 for g(x) in $g(x) = 5^x + k$

$g(x) = 5^x + k$

$-6 = 5^0 + k$

$-6 = 1 + k$

$k + 1 = -6$

$k = -6 - 1$

$k = -7$

For (1,-2) --- Substitute 1 for x and -2 for g(x) in $g(x) = 5^x + k$

$g(x) = 5^x + k$

$-2 = 5^1 + k$

$-2 = 5 + k$

$k+5 = -2$

$k= -2-5$

$k = -7$

To further show that k = -7, we have the following:

$f(x) = 5^x$

When translated downward by 7 units, the function becomes

$g(x) = f(x) - 7$

Recall that: $g(x) = 5^x + k$ and $f(x) = 5^x$

So, we have:

$5^x + k = 5^x - 7$

Subtract $5^x$ from both sides

$5^x - 5^x + k = 5^x - 5^x + 7$

$k = -7$

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