Answer
The equation of the tangent line is $y=-6x+5$
Work Step by Step
$y=-3x^{2}+2$ $;$ $a=1$
First, evaluate the derivative of the given expression:
$y'=-3(2)x=-6x$
Substitute $x$ by $a=1$ in the derivative found to obtain the slope of the tangent line at the given point:
$m_{tan}=-6(1)=-6$
Substitute $x$ by $a=1$ in the original expression to obtain the $y$-coordinate of the point given:
$y=-3(1)^{2}+2=-3+2=-1$
The point is $(1,-1)$
The slope of the tangent line and a point through which it passes are now known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point:
$y-(-1)=-6(x-1)$
$y+1=-6x+6$
$y=-6x+6-1$
$y=-6x+5$
The graph of both the function and the tangent line are shown in the answer section.