If (-1, y) lies on the graph of y = 3^x+1, then what is y?

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If (-1, y) lies on the graph of y = 3^x+1, then what is y?

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Answer 1 · 2015-12-09T06:36:43

it depend on what the original question is
if $y = 3^{x} + 1$ $y= 3^x + 1$ , then solution is $(- 1, \frac{4}{3})$ $(-1, 4/3)$

If $y = 3^{x + 1}$ $y= 3^(x+1)$ , then solution is $(- 1, 1)$ $(-1, 1)$

Explanation:

Assuming your question:
If the point $(- 1, y)$ $(-1, y)$ lies on the graph $y = 3^{x} + 1$ $y= 3^x + 1$ then what is $y$ $y$

We are given $x = - 1$ $x= -1$ , and the $y = 3^{x} + 1$ $y= 3^x +1$ ; let substitute $- 1$ $-1$ for $x$ $x$ into the equation

$y = 3^{- 1} + 1$ $y= 3^(-1) + 1$
$y = \frac{1}{3} + 1$ $y = 1/3 + 1$
$y = \frac{1}{3} + \frac{3}{3}$ $y= 1/3 + 3/3$
$y = \frac{4}{3}$ $y= 4/3$

The point is $(- 1, \frac{4}{3})$ $(-1, 4/3)$

=================
Assuming your question is this
If the point $(- 1, y)$ $(-1, y)$ lies on the graph of $y = 3^{x + 1}$ $y= 3^(x+1)$ , then what is $y$ $y$ ?

Substitute $- 1$ $-1$ for $x$ $x$ into the given equation

$y = 3^{- 1 + 1}$ $y = 3^(-1+1)$
$y = 3^{0}$ $y= 3^0$
$y = 1$ $y= 1$

The point is $(- 1, 1)$ $(-1, 1)$

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If (-1, y) lies on the graph of y = 3^x+1, then what is y?

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