How to draw slope of tangent?

Question

I'm trying to get a sense of what a tangent slope means on a graph. I'll come with a clear example.

slope is $\frac{-1}{x}$

Slope for $(1,1)$, is $-1$

How do I draw the slope? First, I draw the $1$ on $x$ and $1$ on $y$. Then I have a line from $x$ to the height of $1$ on $y$ and join these points. Now where is the slope?

Answer 1

You draw the point $(1,1)$, which looks like

and then draw a line through that point with slope -1, which is diagonal top-left to bottom right, with 1 step down for $y$ for each 1 step to the right for $x$ and looks like

If you wanted to make things clearer, you might sketch part of the curve as well, noting that it gets flatter as $x$ increases and looks like

Answer 2

You have one point on the line that is the slope, $(1,-1)$ (note the minus sign). The equation of a line in point-slope form is $y-y_1=m(x-x_1)$ so plug in your point and slope, getting $y+1=-(x-1)$ or $y=-x$, which you can plot. A slope of $-1$ is a line downward at $45^{\circ}$ to the right.

Answer 3

Here is an attempted explanation. Unfortunately, this answer does not come with pictures, so you will have to draw them for me.

We have the point $P$, say $(1,1)$. We want to draw a line through $P$ with a given slope. I will give somewhat different instructions depending on whether the slope we are interested in is positive or negative or zero. Slope $0$ is very simple. Just draw the line through $P$ parallel to the $x$-axis.

Positive slope: Let the slope we want be say $\frac{2}{3}$. Go to the right of $P$ by $3$ units, then up by $2$ units. Let $Q$ be the point you reach. In our case, $Q=(4,3)$. The line through $P$ and $Q$ has slope $\frac{2}{3}$.

Let's check with algebra. The line through $(x_1,y_1)$ and $(x_2,y_2)$ has slope $\frac{y_2-y_1}{x_2-x_1}$ (the change in $y$ divided by the change in $x$.) So the line through $(1,1)$ and $(4,3)$ has slope $\frac{3-1}{4-1}=\frac{2}{3}$. That's what we wanted. For the rest of this post, we will concentrate on the geometry.

Let's practice a bit. Let $P$ be the point $(1,7)$. We want to draw the line through $P$ that has slope $\frac{5}{3}$. Go to the right by $3$. We reach $(4,7)$. Go up by $5$. We reach $Q=(4,12)$. The line $PQ$ has slope $\frac{5}{3}$.

Now again let $P=(1,7)$. We want to draw the line through $P$ that has slope $3$. Same idea, except that we must write the slope $3$ as $\frac{3}{1}$. Now right by $1$, up by $3$. We reach $Q=(2,10)$.

Negative slope: Suppose we want to draw the line through $P=(1,1)$ that has slope $-\frac{2}{3}$. From $P$, go left by $3$, up by $2$. We reach $Q=(-2,3)$. The line $PQ$ has slope $-\frac{2}{3}$.

Now for something fancy. Let $P=(1,5)$. Draw the line through $P$ that has slope $-\sqrt{3}$. Write the slope as $-\frac{\sqrt{3}}{1}$. Remember, left by $1$, up by $\sqrt{3}$. We reach $Q=(0,5+\sqrt{3}$. The line $PQ$ has slope $-\sqrt{3}$.

One needs practice, and a feeling for what lines of various slopes look like. Lines with positive slope go upwards. Lines with small positive slope, like $\frac{1}{10}$ go upwards kind of slowly, at a small angle. Think of a gently rising road. Lines with big positive slope, like $\frac{9}{2}$ go up fast. Think mountain climbing.

Lines with negative slope go downwards. If the slope is not very negative, like $-\frac{1}{7}$, the line goes down kind of slowly. If the slope is large negative, like $-\frac{31}{3}$, the line goes down steeply.