exponential graphing with intercept

Question

I'm not sure what to do exactly. I got the equation of the line to be $y=\dfrac{p}{\ln\left(p\right)}x+p$ is this correct?

Now I need to solve for x using the above.

Answer 1

Find $z$ such that $f(z)=0$:

$$-e^{-z}+p=0 \Leftrightarrow -e^{-z}=-p \Leftrightarrow e^{-z}=p \Leftrightarrow -z = \ln(p) \Leftrightarrow z= -\ln(p).$$

Compute $f(0)$:

$$f(0)=-e^{-0}+p=-1+p=p-1.$$

Your linear equation needs to go through the points

$$ \begin{aligned} (x_1,y_1)&=(z,0)=(-\ln(p),0)\\ (x_2,y_2)&=(0,f(0))=(0,p-1) \end{aligned}$$

Therefore your linear equation $y(x)=mx+c$ needs to satisfy

$$ \begin{aligned} y(z)&= 0\\ y(0)&= p-1 \end{aligned}$$

You find $c$ by calculating

$$ \begin{aligned} p-1&=y(0)= m\cdot 0 + c = c\\ \end{aligned}$$

After that you can find $m$ by calculating

$$ \begin{aligned} 0=y(z)= m \cdot z + c = m \cdot (-\ln(p)) + p-1 &\Leftrightarrow -(p-1) = m \cdot (-\ln(p))\\ & \Leftrightarrow \frac{p-1}{\ln(p)} = m \end{aligned} $$

You end up with

$$c=p-1 \quad \text{and}\quad m=\frac{p-1}{\ln(p)}.$$

You are searching for the equation

$$y(x)=\frac{p-1}{\ln(p)}x+(p-1).$$

Answer 2

First, you need to find the $x$-intercept of $f(x)$. To do that, you would need to set the function equal to zero, then solve for $x$. The setup would look like this:

$0=-e^{-x}+p\\ p=e^{-x}\\ \ln(p)=-x\\ x=-\ln(p)$

x-intercept = $(-\ln(p),0)$

To get the $y$-intercept, you need to plug in zero for the $x$-value, then solve for $y$. The setup would look like this:

$y=-e^{0}+p\\ y=p-1$

y-intercept = $(0,p-1)$

Now that we have the coordinate points of the $x$ and $y$-intercepts, you need to figure out the equation for a straight line that would pass through both of those points. Since you already know what the $y$-intercept should be, all you need is the slope of the line, which is $\frac{\text{rise}}{\text{run}}$:

slope = $\frac{p-1}{\ln(p)}$

Final equation: $y=\frac{p-1}{\ln(p)}x+(p-1)$