Question

Set up an equation of a tangent to the graph of the following function.
Find the point on the curve $y=4x^2-6x+1$ at which the tangent is parallel to the straight line $y=2x.$

Answer 1

The slope of tangent to curve $y=4x^2-6x+1$ parallel to $y=2x$ is ${}^{\prime }2{}^{\prime }$

$\frac{dy}{dx}=8x-6=slope=2\Rightarrow x=1$

If $x_1=1then{y}_1={4.1}^2-6.1+1=-1$

The equation of line passing through $(1,-1)$ and having slope $m=2$

$(x-1)2=y+1$

$2x-y-3=0$ is the tangent parallel to $y=2x$