Question 1191024
{{{f(x)=x^4-17x^2+16}}}; {{{x=1}}}

Find the first derivative and evaluate at {{{x=1}}} to find a slope of the tangent line.
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{{{f}}}'{{{(x)=4x^3-34x}}}


{{{f}}}'{{{(x)=4*1^3-34*1=-30}}}


Now that we have a slope for the tangent line, we need to identify a point on the line.

We know the tangent line touches the function {{{f(x)}}} at the point {{{x=1}}}, so let's find the value of {{{f(x)}}} at this point:


{{{f(1)=1^4-17*1^2+16=0}}}


So we know the tangent line goes through the point({{{1}}},{{{0}}})


Finally, we can use the point-slope formula for a line to find the equation of the tangent line.

{{{y=mx+b}}}

To find the value of {{{b}}}, substitute the values we have calculated for the point and slope of the tangent line:

{{{0=-30*1+b}}}

{{{b=30}}}


the equation of the tangent line is:

{{{y=-30x+30}}}



{{{ drawing( 600, 600, -15, 15, -55, 55,circle(1,0,.12), locate(1,o.5,p(1,0)),
graph( 600, 600, -15, 15, -55, 55, x^4-17x^2+16,-30x+30)) }}}