Question

Oblicz pole trójkąta wpisanego w funkcję:
y=-x²+8x+7
pilne!

Answer 1

Y=-x²+8x+7
Δ=64-4*(-1)*7=64+28=92
√Δ=√23*4=2√23
x1=-8-2√23/-2=4+√23
x2=-8+2√23/-2=4-√23

y1=-x²+8x+7
y1=-(4+√23)²+8*(4+√23)+7=-(16+8√23+23)+32+8√23=
=-16-8√23-23+32+8√23=
=-39+32=-7
y2=-x²+8x+7
y2=-(4-√23)²+8*(4-√23)+7=-(16-8√23-23)+32-8√23=
=-16+8√23+23+32-8√23=
=39

pole trojkata1=1/2a*h
pole trojkata1=1/2*x1*y1=1/2*4+√23*(-7)=-28+√23/2

pole trojkata2=1/2a*h
pole trojkata2=1/2*x2*y2=1/2*4-√23*(39)=156-√23/2