SL 5.4—Tangents and normal

The tangent line to the graph of $f$ at $x=2$ is given as $y=7-3x$. The slope of this line is the coefficient of $x$, which is $-3$.

By definition, the slope of the tangent line at a point equals the derivative of the function at that point. Therefore $f^{\prime }(2)=m_{\text{tan}}=-3$.

Since the tangent line has slope $-3$, we conclude $f^{\prime }(2)=-3$.

We need the value of $f^{\prime }(2)$, which equals the slope of the tangent at $x=2$. The tangent line equation is $y=7-3x$.

The option $3$ would be the magnitude of the slope but ignores the negative sign. Since the line is $y=7-3x$, the slope is clearly $-3$, not $3$.

The option $1$ has no basis here — it is not the slope, the $y$-intercept, or any other relevant value from the tangent equation.

The option $7$ is the $y$-intercept of the tangent line, not its slope. Confusing the $y$-intercept with the derivative is a common error.

The slope of $y=7-3x$ is $-3$, so $f^{\prime }(2)=-3$.

If the line $y=5x-4$ is tangent to $f$ at $x=1$, then the point of tangency lies on both the curve and the line.

Since the point of tangency has $x=1$, we substitute into the tangent line equation: $y=5(1)-4=1$.

Therefore $f(1)=1$, since the curve and tangent share the same point at $x=1$.

We need $f(1)$, the $y$-coordinate of the curve at $x=1$. At the point of tangency, the curve and tangent line share the same point.

The value $5$ is the slope of the tangent line, i.e. $f^{\prime }(1)=5$, not $f(1)$. Confusing the slope with the function value is a classic error.

The value $-4$ is the $y$-intercept of the tangent line (the constant term), not the $y$-value at $x=1$.

The value $4$ might arise from computing $5(1)+4-4+4$, but this is an arithmetic error. Substituting $x=1$ into $y=5x-4$ gives $y=1$, not $4$.

Substituting $x=1$ into $y=5x-4$ gives $y=5(1)-4=1$, so $f(1)=1$.