Question 1137357


1) The line {{{y= 3x + 4}}} is a tangent to the curve {{{y=ax^3+bx^2 +2x+7 }}}at the point ({{{1}}},{{{7}}}) 

1a.) Determine the values of {{{a}}} and {{{b }}}

use given point ({{{1}}},{{{7}}})  and plug coordinates in {{{y=ax^3+bx^2 +2x+7 }}}
in this case, the curve must go through ({{{1}}},{{{7}}}) 

{{{7=a*1^3+b*1^2 +2*1+7}}}
{{{7=a+b +2+7}}}
{{{a+b=7-9}}}
{{{a+b=-2}}}....solve for {{{a}}}

{{{a=-2-b}}}.....eq.1


and the derivative of {{{y=ax^3+bx^2 +2x+7}}} is 
{{{y}}}'={{{3ax^2+2bx +2}}}

at this point   must be {{{3}}} (the slope of the tangent): 

{{{3a + 2b + 2 = 3}}}
{{{3a + 2b=3- 2 }}}...solve for {{{a}}}
{{{3a =1-2b }}}

{{{a =1/3-2b/3 }}}.......eq.2

from eq.1 and eq.2 we have

{{{-2-b=1/3-2b/3}}}...solve for {{{b}}}

{{{2b/3-b=1/3+2}}}....both sides multiply by {{{3}}}

{{{2b-3b=1+6}}}

{{{-b=7}}}

{{{b=-7}}}

go to eq.1

{{{a=-2-b}}}.....eq.1 substitute {{{b}}}

{{{a=-2-(-7)}}}

{{{a=-2+7}}}

{{{a= 5}}}

=>
{{{y=5x^3-7x^2 +2x+7 }}}


1b. ) The tangent to the curve is parallel to {{{y= 3x +4}}} at one additional point. Determine the x-coordinate of this point. 

For part b, you will need to solve a quadratic using the values of a and b to find the point where the derivative is again = {{{3}}}.

{{{y}}}'={{{3*5x^2+2(-7)x +2}}}
{{{3=15x^2-14x +2}}}
{{{15x^2-14x +2-3=0}}}
{{{15x^2-14x -1=0}}}
{{{15x^2-15x+x -1=0}}}
{{{15x(x-1)+(x -1)=0}}}
{{{(x - 1) (15x + 1) = 0}}}

{{{x=1}}}-> already given in point ({{{1}}},{{{7}}}) 

{{{ 15x + 1= 0}}}->{{{x=-1/15}}}->the x-coordinate of this point