Given the equation:
[tex]y=b\log _3(x+a)[/tex]
Let's find the values of a and b using the given graph.
Since the curve faces up, the value of b will be less than 1.
We have:
Vertical Asymptote at x = -4
Equate to zero:
x + 4 = -4 + 4
x + 4 = 0
Thus, we have:
[tex]y=b\log _3(x+4)[/tex]
Now, take the point:
(x, y) ==> (5, -4)
Input the values for x and y, then solve for b:
Where:
x = 5
y = -4
[tex]\begin{gathered} -4=b\log _3(5+4) \\ \\ -4=b\log _3(9) \\ \\ b=\frac{-4}{\log _39} \\ \\ b=\frac{-4}{\log _3(3^2)} \\ \\ b=\frac{-4}{2\log _33} \\ \\ b=\frac{-2}{\log _33} \\ \\ b=-\frac{2}{1} \\ \\ b=-2 \end{gathered}[/tex]
Therefore, the equation is:
[tex]y=-2\log _3(x+4)[/tex]
• The value for a is , 4
,
• The value for b is , -2