Graph exponential function

Question

I am having problems understanding why $xe^x + 10e^x$ has two $(x,y)$ intercepts. I understand why there is one $(0,10)$, but am unclear on how to return $(-10,0)$.

Any help would be much appreciated, Thanks!

Answer 1

A function like $x\mathrm{e}^x+10\mathrm{e}^x$ does not have an $(x,y)$ intercept. The graph $y=x\mathrm{e}^x+10\mathrm{e}^x$ can have an $x$-intercept, where it crosses the $x$-axis; and it can have a $y$-intercept, where it crosses the $y$-axis.

Consider the equation $y=x\mathrm{e}^x+10\mathrm{e}^x$. There is a common factor of $\mathrm{e}^x$, and so $y=(x+10)\mathrm{e}^x$.

The $y$-intercept comes from putting $x=0$ and then solving for $y$. If $x=0$ then $$y=(0+10)\mathrm{e}^0 = 10$$ There is a single $y$-interecpt with $y$-coordinate $10$

The $x$-intercept comes from putting $y=0$ and solving for $x$. If $y=0$ then $$0=(x+10)\mathrm{e}^x \iff x+10=0 \ \ \text{or} \ \ \mathrm{e}^x=0$$ Since $\mathrm{e}^x>0$ for all real $x$, the only $x$-intercept comes from $x+10=0$, i.e. $x=-10$.

There is a single $x$-interecpt with $x$-coordinate $-10$

The graph $y=(x+10)\mathrm{e}^x$ meets the $y$-axis at $(0,10)$ and meets the $x$-axis at $(-10,0)$.