Question

Set up an equation of a tangent to the graph of the following function.
$y=-x^2-2$ which is parallel to the straight line $y=4x+1.$

Answer 1

Slope of line $y=4x+1$ is ${}^{\prime }{4}^{\prime }$.The slope of a tangent parallel to this line is also ${}^{\prime }{4}^{\prime }$

$y=-x^2-2$

$\frac{dy}{dx}=-2x=4\Rightarrow x=-2$

If $x_1=-2$ then $y_1=-(-2{)}^2-2=-6$

Equation of line passing through $(-2,-6)$ and having slope ${}^{\prime }{4}^{\prime }$ is $(x+2)4=y+6$

$4x-y+2=0$ is the tangent parallel to $y=4x+1$