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Let f be defined by f(x)=−x^2−3x+5. f'(x)=-2x-3 An equation for the tangent line to y=f(x) at x=−1 is y=?
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i guess y comes out to be y=6-x
you know the answer eq?
how did u come out with y=6-x?
at x=-1 y ll be y=7 right? like y= -(-1)^2-3(-1)+5=7
and f'(x)=-1
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putting values in tangent eq (y-7)=(-1)(x-(-1))
y=6-x.isn't it?
that's right, but how did u know that t m=-1?
o thats easy look slope has a point on function and they gave us a point x=-1 ok? put x=-1 in the eq f'(x) =-2x-3 when we put x=-1...f'(x)= -2(-1)-3=2-3=-1
ooh i get it thank u
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welcome:)
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