Question
The Secant Graph has the local minimum point at $(-0.75,-1.5)$ and it has a local maximum at $(2.25,-2.5)$
Give the equation in the form: $y=A \sec(Bx+C)+D$
My thought process&steps until I got stuck
First I find the Period
I do this by finding the horizontal distance from the minimum and maximum and multiply that number by $2$ to get the distance aka the period. that means the distance from the minimum to the maximum is $3$ so the period is $6$.
Then I find $B$
As anyone knows, the way you get the period is to do $\frac{2\pi}{B}$ that means $\frac{2\pi}{B}=6$. then we can deduce that $B$ is going to be $\frac{\pi}{3}$. We also recognize the fact that ($\frac{\pi}{3}(x)+c)=0$ sets up the beginning of the primary period.
Then I find the amplitude
This fairly easy, i just found it to be $0.5$
What I have so far
we have the following right now: $\frac{1}{2}\sec(\frac{\pi}{3}x+C)-2$
Now my question is how to find what is $C$?
After that i plug in -0.75 into the equation as follows:
$\frac{1}{2}\sec(\frac{\pi}{3}(\frac{-3}{4})+C)-2=1.5$
$y=\frac12\sec(\frac\pi3x)-2$ has a minimum at $(0,-1.5)$ and a maximum at $(3, -2.5)$. You only need to shift your graph to the left by $0.75$. Hint: To shift a graph to the left by $C$, change $f(x)$ to $f(x+C)$.