I’m revising my fundamentals on tangents and normals and I’m tripping over the basics. I love how a tangent line is like a “zoomed-in” version of the curve, but when I try to write the actual equation, I keep second-guessing myself.
Example I’m practicing with: y = x^2 − 4x + 1 at x = 2. My (probably wrong) attempt: I plugged x = 2 into the original equation to get y = −3, and then I just used that y-value as the slope. So I wrote the tangent as y = −3x + b, used the point (2, −3) to find b = 3, and got y = −3x + 3. For the normal, I figured it’s “normal” so maybe it keeps the same slope, so I also wrote y = −3x − 3 through the point (2, −3). This feels very off, but I can’t spot why.
What’s the correct, systematic way to go from the curve to:
– the slope of the tangent at x = 2,
– the tangent line equation,
– and then the normal line?
I know derivatives are involved, but where exactly do they enter, and how do I handle special cases (e.g., if the tangent is horizontal or vertical)? Also, I always mix up reciprocal vs negative reciprocal for normals-any memory trick?
I’m trying to strengthen my fundamentals here, so a clear step-by-step would help me lock in the pattern. Any help appreciated!
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3 Responses
I think the systematic steps are: take the derivative for the slope (not the y-value), so f'(x)=2x−4 gives slope 0 at x=2 and point (2,−3), hence the tangent is y=−3 (horizontal) and the normal is x=2 (vertical), using the template y−f(a)=f'(a)(x−a).
Special cases and memory trick: if f'(a)=0 the normal is vertical, if the tangent is vertical the normal is horizontal, otherwise m_normal = −1/m_tangent because perpendicular slopes multiply to −1; quick refresher: https://tutorial.math.lamar.edu/classes/calci/tangentsnormals.aspx.
For y = x^2 − 4x + 1 at x=2, y’ = 2x − 4 gives m_tan = 0 at (2, −3), so the tangent is y = −3 (horizontal) and the normal is x = 2 (vertical). In general, m_tan = y'(x0) and y − y0 = m_tan(x − x0); the normal has slope −1/m_tan when m_tan ≠ 0 (perpendicular slopes multiply to −1), with m_tan = 0 ⇒ vertical normal and m_tan undefined ⇒ horizontal normal-see https://tutorial.math.lamar.edu/classes/calci/tangentline.aspx; Hope this helps!
Your intuition about “zooming in” is right, but the slope of the tangent comes from the derivative, not the y-value. For f(x) = x^2 − 4x + 1 at x = 2, first find the point: f(2) = −3, so the point is (2, −3). Differentiate: f′(x) = 2x − 4, hence f′(2) = 0. That means the tangent is horizontal, so its equation is y = −3 (from y − (−3) = 0·(x − 2)). The normal is perpendicular to the tangent: in general, if the tangent slope is m, the normal slope is −1/m (their product is −1). Here m = 0, so the normal is vertical, with equation x = 2. Memory trick: “perpendicular = flip and switch sign” (negative reciprocal); treat m = 0 as flipping to “infinite slope” (vertical line), and if the tangent were vertical (derivative undefined), the normal would be horizontal y = f(a). Small analogy: the tangent is the direction your car points on the road at that instant (its steepness is the derivative), while the normal is the fencepost sticking straight out from the roadside.