The equation of a line in the plane is
#y=mx+q#
When we have a function #f(x)#, the line tangent to the function in a specific point has the #m# that is the derivative of the function in that point. That is #m=(df(p_0))/dx# where #p_0# is the desired point.
So first of all we have to calculate the derivative
#(df)/dx=14x-3#
then, for #x=1#, we have
#m=(df(1))/dx=11#.
Our line is then
#y=11x+q#.
We still miss #q#, but it is enough to know that the line passes for one point to evaluate it. We know that the line is tangent to the function in #x=1#, so we are sure that our line will pass from the point
#x=1, y=f(1)=7*1^2-3*1+6=10#.
We substitute this point in the equation of the line
#10=11*1+q# obtaining #q=-1#.
Then the equation of the line is
#y=11x-1#.
