Question

Given A(-2,1) and B(3,-4)

Answer 1

$\bf f(x)=-2x^2-5x-7\\\\ -------------------------------\\\\ \cfrac{f(x+h)-f(x)}{h} \\\\\\ \cfrac{[-2(x+h)^2-5(x+h)-7]~~-~~[-2x^2-5x-7]}{h} \\\\\\ \cfrac{[-2(x^2+2xh+h^2)-5(x+h)-7]~~-~~[-2x^2-5x-7]}{h} \\\\\\ \cfrac{\underline{-2x^2}-4xh-2h^2\underline{-5x}-5h\underline{-7}\underline{+2x^2+5x+7}}{h} \\\\\\ \cfrac{-4xh-2h^2-5h}{h}\implies \cfrac{\underline{h}(-4x-2h-5)}{\underline{h}}\implies -4x-2h-5$

$\bf -------------------------------\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad B(\stackrel{x_2}{3}~,~\stackrel{y_2}{-4})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AB=\sqrt{[3-(-2)]^2+[-4-1]^2}\implies AB=\sqrt{(3+2)^2+(-4-1)^2} \\\\\\ AB=\sqrt{25+25}\implies AB=\sqrt{2\cdot 25} \\\\\\ AB=\sqrt{2\cdot 5^2}\implies AB=5\sqrt{2}\\\\ -------------------------------$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad B(\stackrel{x_2}{3}~,~\stackrel{y_2}{-4})\qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{3-2}{2}~~,~~\cfrac{-4+1}{2} \right)\implies \left(\frac{1}{2}~~,~~-\frac{3}{2} \right)\implies \left(\frac{1}{2}~~,~~-1\frac{1}{2} \right)$