Tangent lines & slopes

1. Identify the point of tangency $(a,f(a))$:

The point of tangency is where the tangent line meets the curve. This is the function value:

$$f(-3)=(-3{)}^2=9$$

2. Find the slope of the tangent line $f^{\prime }(a)$:

In chapter 2.1.1, the derivative of $f(x)=x^2$ was found to be

$$f^{\prime }(x)=2x$$

The slope of the tangent line at $x=-3$ is the derivative evaluated at the point:

$$f^{\prime }(-3)=2(-3)=-6$$

3. Construct the equation of the tangent line:

The tangent line meets the curve at $(-3,9)$ and has a slope of $-6$, so its equation is:

$$y-9=-6(x+3)$$

or in slope-intercept form,

$$y=-6x-9.$$

It helps to plot both $y=x^2$ and $y=-6x-9$ in Desmos to see what a tangent line looks like relative to the curve.

1. Find the slope of the given line:

First, rewrite the line $2x+y=6$ in slope-intercept form:

$$y=-2x+6$$

This equation shows the slope of the line is $-2$. Since parallel lines have the same slope, the tangent line to $f$ that we’re looking also has a slope of $-2$.

2. Find the point of tangency $(a,f(a))$:

At the target point $x=a$, the tangent line to $f$ has slope $-2$. In other words, $f^{\prime }(a)=-2$. We just need to find $a$.

Since the derivative of $f(x)=x^2$ is $f^{\prime }(x)=2x$, we solve:

$$\begin{array}{rl}{f}^{\prime }(a)& =2a\\ -2& =2a\\ a& =-1\end{array}$$

Then the corresponding $y$-value is

$$f(-1)=(-1{)}^2=1$$

So the tangent line to $f$ meets it at $(-1,1)$.

3. Construct the equation of the tangent line:

The tangent line passes through $(-1,1)$ and has a slope of $-2$, so its equation is

$$y-1=-2(x+1)$$

or in slope-intercept form,

$$y=-2x-1.$$

1. Identify $(a,f(a))$:

The function value at $x=4$ is

$$f(4)=4^2=16$$

2. Find the slope of the normal line:

The derivative of $x^2$ is $f^{\prime }(x)=2x$. Evaluate at $x=4$ for the slope of the tangent line:

$$f^{\prime }(4)=2(4)=8$$

The slope of the normal line is the negative reciprocal, or $-\frac{1}{8}$.

3. Construct the equation of the normal line:

Since the normal line to $f$ at the point $(4,16)$ has a slope of $-\frac{1}{8}$, its equation is

$$y-16=-\frac{1}{8}(x-4)$$

It helps to confirm visually by plotting the original function $f(x)=x^2$, the equation of the tangent line, and the equation of the normal line.

The instantaneous rate of change of the height means the derivative, $h^{\prime }(t)$. Our goal is to find the time $t$ such that $h^{\prime }(t)=0$.

Below is the derivative using the limit definition. To keep the notation clear, the height function is rewritten as $f(x)=-x^2+4x$.

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{[-(x+h{)}^2+4(x+h)]-(-x^2+4x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{-(x^2+2xh+h^2)+4x+4h+x^2-4x}{h}\\ & =\underset{h\to 0}{\lim }\frac{-2xh-h^2+4h}{h}\\ & =\underset{h\to 0}{\lim }\frac{\cancel{h}(-2x-h+4)}{\cancel{h}}\\ & =-2x+4\end{array}$$

Setting this equal to $0$:

$$\begin{gathered} -2x+4=0 \\ x=2 \end{gathered}$$

Therefore, the instantaneous rate of change of the ball’s height is $0$ when $t=2$.

Observing the graph of the parabola, this is the highest point the ball reaches (the vertex), and the slope of the tangent line to the curve is $0$.

It’s highly recommended that you sketch a quick graph of $y=x^2$. Try drawing the tangent line that passes through $(0,-1)$ with a negative slope to see where it might meet the curve.

1. Identify the point of tangency:

The tangent line to $y=x^2$ meets it at the point $(a,a^2)$.

2. Slope of the line

The slope $m$ can be expressed in two ways:

1. Using algebra:

Since the line passes through $(0,-1)$ and $(a,a^2)$, the slope can be expressed in terms of $a$ as:

$$\begin{array}{rl}\text{Slope}& =\frac{y_2-y_1}{x_2-x_1}\\ m& =\frac{a^2+1}{a}\end{array}$$

2. Using calculus:

The slope of the tangent line at any point on $y=x^2$ is the derivative. Since $f^{\prime }(x)=2x$, and $m$ is the slope at $x=a$, then

$$m=2a$$

3. Solve for $a$:

Set the two expressions for $m$ equal to each other.

$$\begin{array}{rl}2a& =\frac{a^2+1}{a}\\ 2a^2& =a^2+1\\ a^2& =1\\ a& =\pm 1\end{array}$$

Since we want the line with a negative slope, take $a=-1$, which gives

$$m=-2$$

Note that if the question asked for the positive slope ($m>0$), we would take $a=1$ such that $m=2(1)=2$. This second tangent line also passes through $(0,-1)$ but meets $y=x^2$ at $(1,1)$ instead.