Answer
- Transformation 1 has no y-intercept
- Transformation 2 has a y-intercept of -2
SOLUTION
Problem Statement
We are told the PH of a pool is given by the function:
[tex]p(t)=-\log 10t[/tex]New chemicals transform this function and we are asked to find which transformation results in a y-intercept and we are also asked to show them graphically.
The translations are:
[tex]\begin{gathered} p(t)+1=f(t)=-\text{log(}t)+1\text{ (Transformation 1)} \\ P(t+1)-1=f(t)=-\log 10(t+1)-1\text{ (Transformation 2)} \end{gathered}[/tex]Method
Y-intercept
To find the y-intercepts of the graphs, we simply need to find the value of the p(t) when t = 0.
Plotting the graphs:
To plot the graphs, we shall apply a graphing calculator.
Solution
Transformation 1:
i. Y-intercept:
[tex]\begin{gathered} f(t)=-\log (t)+1 \\ put\text{ t=0} \\ f(0)=-\log (0)+1 \\ \\ \text{But we know that,} \\ \lim _{t\to0}\log (t)\to-\infty \\ \therefore\lim _{t\to0}-\log (t)\to\infty \\ \\ \therefore\lim _{t\to0}f(t)=\lim _{t\to0}(-\log (t)+1)\to\infty \end{gathered}[/tex]Thus, we can see that the y-intercept of f(t) tends towards infinity and since infinity is not on the Real Number line, we can conclude that this transformation has no y-intercept
ii. Plotting Transformation 1:
The graph of Transformation 1 is given below:
Transformation 2:
i. Y-intercept:
[tex]\begin{gathered} f(t)=-\log 10(t+1)-1 \\ \text{put t=0} \\ f(0)=-\log 10(0+1)-1 \\ f(0)=-\log 10-1 \\ \\ \text{ We know that }log10=1 \\ \therefore f(0)=-(1)-1=-1-1 \\ \\ f(0)=-2 \end{gathered}[/tex]Thus, the y-intercept of this transformation exists and it is -2.
ii. Plotting Transformation 2
Final Answer
- Transformation 1 has no y-intercept
- Transformation 2 has a y-intercept of -2
