SL 5.4—Tangents and normals

$f^{\prime }(x)=2x-5$.

$f^{\prime }(3)=2(3)-5=6-5$.

$m_T=1$.

The slope of the tangent at $x=3$ is $1$.

$-1$ comes from a sign error when computing $6-5$.

$6$ is $f^{\prime }(3)$ before subtracting 5, i.e. forgetting the constant term of the derivative.

$5$ mistakenly treats the coefficient $-5$ alone as the answer.

Only careful substitution gives $m_T=1$.

$\frac{dy}{dx}=3x^2$.

At $x=1$, $m_T=3(1{)}^2=3$.

$m_N=-\frac{1}{m_T}=-\frac{1}{3}$.

The normal slope is $-\frac{1}{3}$.

$3$ is the tangent slope, not the normal slope.

$-3$ takes the negative but forgets to reciprocate.

$\frac{1}{3}$ reciprocates but forgets the negative sign.

Only the negative reciprocal, $-\frac{1}{3}$, is correct.