From a Tangent Slope to a Line
A tangent slope describes how steep a curve is at a particular point. Once that slope and the point of contact are known, a line equation can be constructed using the same information used to describe any line: a point on the line and its slope.
This tutorial focuses on that construction. The slope is treated as known—for example, because it has been found by examining secant slopes as their interval shrinks. The task is to turn the slope and contact point into an equation, then check that the equation gives the right point.
In this definition, the nearby point has the form \((x,f(x))\), with \(x\ne a\). The secant slope is \(\frac{f(x)-f(a)}{x-a}\). For a two-sided limiting tangent slope, this quotient must approach the same finite value \(m\) as \(x\) approaches \(a\) from either side. Here we will use the value of \(m\) once it is known, rather than finding it from scratch.
Point-Slope Form
A line with slope \(m\) passing through a known point \((x_1,y_1)\) can be written in point-slope form:
For a tangent line, the contact point supplies the coordinates \(x_1=a\) and \(y_1=f(a)\). The limiting tangent slope supplies \(m\). Substituting those three values gives the tangent-line equation:
The symbols have distinct roles. The value \(a\) is the input at the contact point, \(f(a)\) is the corresponding output, and \(m\) is the slope. In particular, the point is \((a,f(a))\), not \((a,m)\): the slope is not generally the curve’s output.
Why the Formula Works
The point-slope equation follows directly from what slope means. Take any point \((x,y)\) on a line through \((a,f(a))\). When \(x\ne a\), the line’s slope is the change in the vertical coordinate divided by the change in the horizontal coordinate:
For the line to have slope \(m\), this ratio must equal \(m\). Multiplying by \(x-a\) gives \(y-f(a)=m(x-a)\). The equation also includes the contact point: if \(x=a\), then its right-hand side is zero, so it requires \(y-f(a)=0\), or \(y=f(a)\). Thus the equation describes a line of slope \(m\) that passes through the curve’s contact point.
A Reliable Construction
When the contact point and limiting slope are given, keep the construction in a fixed order. This helps avoid mixing the coordinates with the slope.
If the contact input is \(a\), evaluate the curve to get \(f(a)\). Record the point as \((a,f(a))\).
Use the stated or previously established value \(m\), including its sign.
Write \(y-f(a)=m(x-a)\) before expanding or rearranging.
Set \(x=a\) in the line equation and confirm that it gives \(y=f(a)\).
Worked Examples
Worked Example: A Positive Tangent Slope
Suppose the curve is \(y=f(x)=x^2\), and its limiting tangent slope at \(x=2\) is known to be \(m=4\). Find the tangent-line equation and check that it passes through the curve’s contact point.
First find the output at the contact input:
The contact point is therefore \((2,4)\). Substitute \(a=2\), \(f(a)=4\), and \(m=4\) into point-slope form:
Expanding gives:
To check the contact point, set \(x=2\) in the line equation. It gives \(y=4(2)-4=8-4=4\), which agrees with \(f(2)=4\). The line \(y=4x-4\) passes through \((2,4)\).
Worked Example: A Negative Tangent Slope
Consider \(y=f(x)=x^2\) at \(x=-1\), where the limiting tangent slope is known to be \(m=-2\). Find the tangent line.
The curve’s output at the contact input is:
Thus the contact point is \((-1,1)\). Substitute \(a=-1\), \(f(a)=1\), and \(m=-2\):
Expand and solve for \(y\):
For the check, substitute \(x=-1\) into the line: \(y=-2(-1)-1=2-1=1\). The curve also gives \(f(-1)=1\), so the line passes through \((-1,1)\). The negative slope is reflected in the line equation’s coefficient of \(x\).
Worked Example: A Horizontal Tangent Line
Suppose \(y=f(x)=3+x^2\), and the limiting tangent slope at \(x=0\) is \(m=0\). Find the tangent line and verify the contact point.
Evaluate the curve at the contact input:
The point of contact is \((0,3)\). Point-slope form gives:
Since \(0(x-0)=0\), the equation reduces to \(y=3\). Checking \(x=0\) in this line gives \(y=3\), exactly the curve value \(f(0)=3\). A line with zero slope is horizontal, and here it passes through the required contact point.
Checking and Interpreting the Equation
For a line to be the tangent line at a stated contact point, it must meet two requirements: it must pass through the contact point, and it must have the given limiting tangent slope. Point-slope form builds in both requirements, but checking them is a useful way to catch substitution or simplification errors.
| Check | What to verify |
|---|---|
| Contact point | Substituting \(x=a\) into the line gives \(y=f(a)\). |
| Slope | In point-slope form, the multiplier of \(x-a\) is \(m\). |
| Rearrangement | If written as \(y=mx+b\), the coefficient of \(x\) remains \(m\). |
Point-slope form is often the clearest final answer because it displays the contact point and slope directly. Slope-intercept form, \(y=mx+b\), can be convenient for graphing or comparing lines, but it is not necessary to rearrange if point-slope form already answers the question.
Common Mistake: Losing the Sign of the Contact Input
Suppose the contact point is \((-1,1)\) and the slope is \(-2\). A common error is to write \(x-1\) instead of \(x-(-1)\), producing the incorrect calculation:
This line does not pass through the stated contact point: substituting \(x=-1\) gives \(y=-2(-1)+3=5\), not \(1\). The correction is to preserve the coordinate \(-1\) in the point-slope expression:
Now substituting \(x=-1\) gives \(y=1\), as required. Writing the point as an ordered pair before substituting helps prevent this sign error.
Keep the Curve and Its Tangent Line Distinct
The curve and the tangent line are different equations. The curve gives the output at every input according to \(y=f(x)\); the tangent line is a straight line determined by one contact point and one limiting slope. The line is required to match the curve’s point and slope at contact, but it is not generally the same as the curve elsewhere.
For that reason, the final check in these examples is local and specific: evaluate the proposed line at the contact input and compare its output with the curve’s output there. This confirms that the equation is anchored at the correct point. The slope in the point-slope formula records the tangent slope already supplied for that point.
Check Your Understanding
For each question, keep the contact point and slope separate until you substitute them into point-slope form.
- A curve has contact point \((3,5)\) and limiting tangent slope \(2\). Write its tangent-line equation in point-slope form.
- The curve \(y=f(x)\) has \(f(1)=-2\), and its limiting tangent slope at \(x=1\) is \(-3\). Find the tangent line and check the contact point.
- Why does substituting \(x=a\) into \(y-f(a)=m(x-a)\) verify that the line contains the contact point?
- A proposed line is \(y-4=3(x+2)\) at a claimed contact point \((-2,4)\). Does it pass through that point? Show the substitution.
- What information does \(m\) represent in \(y-f(a)=m(x-a)\), and what information does \(f(a)\) represent?
