Find an equation of the line that satisfies the given conditions.Through (1/2, -2/7) perpendicular to the line 5x-10y=1
5x−10y=1⇒g(x)=0.5x−0.1f(x)=ax2+bx+c1. f(x)=g(x)and2. f′(x)∗g′(x)=−1
f(x)=ax2+bx+c2=a∗12+b∗1+c7=a∗(−2)2+b(−2)+c(2)2=a+b+c(1)7=4a−2b+c(1)−(2)5=3a−3b⇒b=a−53 and c=113−2a
f(x)=g(x)ax2+bx+c=0.5x−0.1ax2+(a−53)x+113−2a=0.5x−0.1⇒x1,2=13−6a12a±√(6a−13)2(12a)2−(113−60a)30a
f′(x)∗g′(x)=−1f′(x)=2ax+bg′(x)=0.5(2ax+b)∗0.5=−1b=a−53⇒2ax+b=−2⇒2ax+(a−53)=−22ax+a=−13
2a[13−6a12a±√(6a−13)2(12a)2−113−60a30a]+a=−13⇒324a2−542.4a−56=0⇒a1,2=22.6±√636.7627
a must be negative: a=22.6−√636.7627
a=(22.6−√636.76)27⇒a=−0.0975594039708354
b=(22.6−√636.76)27−53⇒b=−1.7642260706375021
c=113−2×((22.6−√636.76)27)⇒c=3.8617854746083375
The eqaution is:
f(x)=ax2+bx+cf(x)=−0.0975594039708354x2−1.7642260706375021x+3.8617854746083375

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