Question 1191024
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At x=1, we have f(1)=1-17+16=0, so the point of tangency is (1,0)<br>
To find the slope at (1,0), evaluate the derivative of the function at x=1.<br>
f'(x)=4x^3-34x
f'(1)=4-34=-30<br>
The slope at x=1 is -30.<br>
Plug (x,y)=(1,0) and slope m=-30 into either the point-slope or slope-intercept form of the equation of a line to find the equation of the tangent is y=-30x+30.<br>
A graph with x on (-2,2) and y on (-20,20):<br>
{{{graph(400,400,-2,2,-20,20,x^4-17x^2+16,-30x+30)}}}<br>