# How do you find the point on the line y=4x + 7 that is closest to the point (0,-3)?

The distance between and arbitrary point $\left(x , y\right) = \left(x , 4 x + 7\right)$ on this line and the point $\left(0 , - 3\right)$ is $\setminus \sqrt{{\left(x - 0\right)}^{2} + {\left(4 x + 7 - \left(- 3\right)\right)}^{2}} = \setminus \sqrt{17 {x}^{2} + 80 x + 100}$.
Minimizing the squared distance will occur at the same value of $x$ where the distance is minimized, therefore, we can focus on minimizing the function $f \left(x\right) = 17 {x}^{2} + 80 x + 100$. Its derivative is $f ' \left(x\right) = 34 x + 80$, which has one root at $x = - \frac{80}{34} = - \frac{40}{17} \setminus \approx - 2.353$.
That this value of $x$ gives a minimum distance is clear since the graph of $f \left(x\right)$ is a parabola opening upward, though you can also note that the second derivative is $f ' ' \left(x\right) = 34 > 0$ for all $x$, making the graph of $f$ concave up.
When $x = - \frac{40}{17}$, $y = 4 \setminus \cdot \left(- \frac{40}{17}\right) + 7 = \setminus \frac{119 - 160}{17} = - \frac{41}{17}$. Hence, the point $\left(x , y\right) = \left(- \frac{40}{17} , - \frac{41}{17}\right)$ is the point on the line $y = 4 x + 7$ that is closest to the point $\left(0 , - 3\right)$. The minimum distance itself is $\setminus \sqrt{f \left(- \frac{40}{17}\right)} = \setminus \sqrt{\frac{100}{17}} = \frac{10}{\sqrt{17}} = \setminus \frac{10 \setminus \sqrt{17}}{17} \setminus \approx 2.425$.