Find the slope of the tangent line to the curve $x = y ^2 - 4y$ at the points where the curve crosses the $y$-axis.

I've never been able to understand word problems in math. What is the question asking? What am I supposed to do with it?

Find the slope of the tangent line to the curve $x = y ^2 - 4y$ at the points where the curve crosses the $y$-axis.

• You seem to be asking lots of elementary calculus questions. That's OK - it's what this site is for. When you get answers that help you, please upvote them, and accept the best. Note: you will get more useful answers if you show how you started a problem and where you got stuck. – Ethan Bolker Mar 25 '16 at 14:07
• I often do, but sometimes I can't understand the problem at all. By the way, is there a way to mark a question as answered? – user6050977 Mar 25 '16 at 14:19
• You accept an answer to your question by clicking the checkmark in the margin. You upvote with the up arrow. See math.stackexchange.com/help/someone-answers and math.stackexchange.com/help/accepted-answer . – Ethan Bolker Mar 25 '16 at 14:24

In basic calculus ( to be focused ), a curve crosses the $\;y -$axis exactly when $\;x=0\;$ (if the function's defined there at all), so in your case it is $\;0=y(y-4)\iff y=0\,,\,\,y=4\;$ and we have two points on the $\;y -$ axis: $\;(0,0)\;,\;\;(0,4)\;$ .
The slope of the tangent line at any point on the curve $\;=\;$ the derivative at that point. In your case, implicitly differentiating:
$$x=y^2-4y\implies 1=2yy'-4y'=(2y-4)y'\implies$$
\begin{align*}&\text{at}\;\;(0,0)\;,\;\;1=(2\cdot0 -4)y'\implies y'=-\frac14\\{}\\&\text{at}\;\;(0,4)\;,\;\;1=(2\cdot4-4)y'\implies y'=\frac14\end{align*}
• You differentiate $\;x\;$ with respect to $\;x\;$ (and thus get $\;1\;$) , and $\;y\;$ also with respect to $\;x\;$ . Google "implicit differentiation". – DonAntonio Mar 25 '16 at 14:12