Question

Set up an equation of a tangent to the graph of the following function.
Find the point on the curve $y=x^2-x+1$ at which the tangent is parallel to the straight line $y=3x-1.$

Answer 1

The slope of tangent to curve $y=x^2-x+1$ parallel to $y=3x-1$ is ${}^{\prime }3{}^{\prime }$

$\frac{dy}{dx}=2x-1=slope=3\Rightarrow x=2$

If $x_1=2then{y}_1=2^2-2+1=3$

The equation of line passing through $(2,3)$ and having slope $m=3$

$(x-2)3=y-3$

$3x-y-3=0$ is the tangent parallel to $y=3x-1$