Question 946753
The slope of the tangent is equal to the value of the derivative at the point.
{{{f(x)=x^2-6x}}}
{{{df/dx=2x-6}}}
So at {{{x=-5}}}
{{{m=2(-5)-6}}}
{{{m=-16}}}
So then,
{{{f(-5)=25+30=55}}}
and
{{{y-55=-16(x-(-5))}}}
{{{y-55=-16(x+5)}}}
{{{y-55=-16x-80}}}
{{{highlight(y=-16x-25)}}}
.
.
.
{{{graph(300,300,-10,10,-10,90,x^2-6x,-16x-25)}}}