The distance between and arbitrary point (x,y)=(x,4x+7)(x,y)=(x,4x+7) on this line and the point (0,-3)(0,−3) is \sqrt{(x-0)^2+(4x+7-(-3))^2}=\sqrt{17x^{2}+80x+100}√(x−0)2+(4x+7−(−3))2=√17x2+80x+100.
Minimizing the squared distance will occur at the same value of xx where the distance is minimized, therefore, we can focus on minimizing the function f(x)=17x^{2}+80x+100f(x)=17x2+80x+100. Its derivative is f'(x)=34x+80, which has one root at x=-80/34=-40/17\approx -2.353.
That this value of x gives a minimum distance is clear since the graph of f(x) is a parabola opening upward, though you can also note that the second derivative is f''(x)=34>0 for all x, making the graph of f concave up.
When x=-40/17, y=4\cdot(-40/17)+7=\frac{119-160}{17}=-41/17. Hence, the point (x,y)=(-40/17,-41/17) is the point on the line y=4x+7 that is closest to the point (0,-3). The minimum distance itself is \sqrt{f(-40/17)}=\sqrt{100/17}=10/sqrt(17)=\frac{10\sqrt{17)}{17} \approx 2.425.
Here's a picture for this situation. The line segment intersects the given line at a right angle.