Tangent Line at a Point

Geometry Level 1

What is the equation of the line that passes through the point $(3,24)$ and is perpendicular to the tangent line of the curve $y=x^2+5x$ at that point?

y = \frac1{11} x + \frac{267}{11}

y = \frac1{11} x - \frac{267}{11}

y = -\frac1{11} x + \frac{267}{11}

y = -\frac1{11} x - \frac{267}{11}

1 solution

Brilliant Mathematics Staff
Jul 31, 2020

The slope of the curve $y=x^2+5x$ at $(3,24)$ is $\begin{aligned} y' &= 2x + 5 \\ \Rightarrow \left. y' \right\vert_{x=3} &= 2\cdot 3 + 5 \\ &= 11, \end{aligned}$ which implies that the slope of the line of interest is $-\frac{1}{11}.$

Thus, our line has a slope of $-\frac{1}{11}$ and passes through $(3,24),$ and therefore its equation is $\begin{aligned} y-24 &= -\frac{1}{11} (x-3) \\ \Rightarrow y &= -\frac{1}{11}x + \frac{267}{11}. \end{aligned}$

Tangent Line at a Point

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