Question

a student claims that the line tangent to the graph of f(x)=e^x at x = 1 passes through the point (0,0). is she correct? will the line tangent at x = 2 pass through (1,0)? explain. is the student correct in saying that the line tangent to the graph of f(x)=e^x at x = 1 passes through the point (0,0)? no yes cannot be determined will the line tangent at x = 2 pass through (1,0)? explain. the line tangent at x = 2 pass through the point (1,0) because the tangent line is the line passing through at (1,0) and (2,f(2)).

Explanation:

Step1: Find the derivative of $y = e^{x}$

The derivative of $y = e^{x}$ is $y'=e^{x}$.

Step2: Find the slope of the tangent - line at $x = 1$

When $x = 1$, the slope $m=e^{1}=e$. Also, when $x = 1$, $y = e^{1}=e$. So the point - slope form of the tangent line is $y - e=e(x - 1)$. Simplify it: $y - e=ex - e$, then $y = ex$. When $x = 0$, $y = 0$. So the tangent line to $y = e^{x}$ at $x = 1$ passes through $(0,0)$.

Step3: Find the slope of the tangent - line at $x = 2$

When $x = 2$, the slope $m = e^{2}$. And when $x = 2$, $y = e^{2}$. The point - slope form of the tangent line is $y - e^{2}=e^{2}(x - 2)$. Expand it: $y - e^{2}=e^{2}x-2e^{2}$, then $y = e^{2}x - e^{2}$.

Step4: Check if the tangent line at $x = 2$ passes through $(1,0)$

Substitute $x = 1$ and $y = 0$ into $y = e^{2}x - e^{2}$. We get $0=e^{2}\times1 - e^{2}=0$. So the tangent line at $x = 2$ passes through $(1,0)$.

Answer:

Is the student correct in saying that the line tangent to the graph of $f(x)=e^{x}$ at $x = 1$ passes through the point $(0,0)$?

• B. Yes

Will the line tangent at $x = 2$ pass through $(1,0)$?

• Yes, because the tangent line equation $y = e^{2}x - e^{2}$ satisfies the point $(1,0)$ when substituting $x = 1$ and $y = 0$.