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CBSE Class 12-science Answered

find the equation of the tangent to the curve y=-5x square+6x+7 at the point (1/2,35/4).
Asked by pragati nargotra | 19 Nov, 2014, 06:23: PM
answered-by-expert Expert Answer
C o n s i d e r space t h e space e q u a t i o n space o f space t h e space c u r v e : y equals minus 5 x squared plus 6 x plus 7 S l o p e equals fraction numerator d y over denominator d x end fraction subscript open parentheses 1 half comma 35 over 4 close parentheses end subscript fraction numerator d y over denominator d x end fraction equals minus 10 x plus 6 rightwards double arrow fraction numerator d y over denominator d x end fraction subscript open parentheses 1 half comma 35 over 4 close parentheses end subscript equals open parentheses minus 10 x plus 6 close parentheses subscript open parentheses 1 half comma 35 over 4 close parentheses end subscript rightwards double arrow fraction numerator d y over denominator d x end fraction subscript open parentheses 1 half comma 35 over 4 close parentheses end subscript equals open parentheses minus 10 cross times 1 half plus 6 close parentheses rightwards double arrow fraction numerator d y over denominator d x end fraction subscript open parentheses 1 half comma 35 over 4 close parentheses end subscript equals open parentheses minus 5 plus 6 close parentheses rightwards double arrow fraction numerator d y over denominator d x end fraction subscript open parentheses 1 half comma 35 over 4 close parentheses end subscript equals 1 E q u a t i o n space o f space t h e space tan g e n t space w i t h space s l o p e space m space a n d space p a s sin g space t h r o u g h space t h e space p o i n t space open parentheses x subscript 1 comma y subscript 1 close parentheses space i s y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses E q u a t i o n space o f space t h e space tan g e n t space w i t h space s l o p e space m equals 1 space a n d space p a s sin g space t h r o u g h space t h e space p o i n t space open parentheses 1 half comma 35 over 4 close parentheses i s y minus 35 over 4 equals 1 cross times open parentheses x minus 1 half close parentheses rightwards double arrow fraction numerator 4 y minus 35 over denominator 4 end fraction equals open parentheses fraction numerator 2 x minus 1 over denominator 2 end fraction close parentheses rightwards double arrow fraction numerator 4 y minus 35 over denominator 2 end fraction equals 2 x minus 1 rightwards double arrow 4 y minus 35 equals 2 open parentheses 2 x minus 1 close parentheses rightwards double arrow 4 y minus 35 equals 4 x minus 2 rightwards double arrow 4 x minus 4 y plus 33 equals 0
Answered by Vimala Ramamurthy | 20 Nov, 2014, 09:31: AM
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