Question 1159785

The line {{{y=2x+c}}} is a tangent to the curve {{{y=2x^2-6x+20 }}} (assuming you have {{{highlight(-6x)}}}) 

find the value of c

{{{f(x) = 2x^2-6x+20 }}}

first derivate:

{{{f}}}'{{{(x) =  4x-6 }}}

For the tangent {{{ 2x+c }}} to just touch {{{f(x)}}}, we need to find where{{{ f(x)}}} has slope equal to {{{2}}}:

{{{ 4x-6 = 2 }}} -> remember, the entire LHS is the slope of {{{ f(x)}}}
{{{ 4x= 2+6 }}}
{{{ 4x= 8 }}}
{{{x = 2  }}}      


At {{{x=2}}}:  

{{{f(2) = 2*(2^2) - 6(2) + 20 = 8-12+20 = 16 }}}

So the tangent line {{{ 2x+c }}} just meets{{{ f(x)}}} at {{{x=2}}}, hence it has value {{{16}}} there:

then
{{{ 2x + c = 2(2) + c}}}
{{{4+c = 16}}}  
{{{ c = 12}}}


your answer is:  

{{{ highlight(c = 12) }}}  and the tangent line is {{{ y=2x+12}}}, and the point of tangency is ({{{2}}},{{{16}}})
    


{{{drawing ( 600, 600, -5, 25, -5, 25,
circle(2,16,.12),locate(2,16,p(2,16)),
graph( 600, 600, -5, 25, -5, 25, 2x+12, 2x^2-6x+20)) }}}