Question

Set up an equation of a tangent to the graph of the following function.
$y=2x^2+1$ at the points (1, 5).

Answer 1

$y=2x^2+1$

$\frac{dy}{dx}=4x$

${\left(\frac{dy}{dx}\right)}_{x=1}$=$4$

$(x_1,y_1)=(1,5)$

The equation of the tangent is $(x-x_1)m=y-y_1$

$4(x-1)=y-5$

$4x-y+1=0$ is the equation of tangent