Answer to Question #176303 in Calculus for jimz lester

Question

find an equation of the line tangent to the curve y=3x²-1 and parallel to the line 2x-y+3

Answer 1

Let us find an equation of the line tangent to the curve $y=3x^2-1$ and parallel to the line $2x-y+3=0.$ The equation of the tangent line at point $x_0$ is $y=y(x_0)+y'(x_0)(x-x_0)$. Since the tangent line and the line $y=2x+3$ are parallel, their slopes coinside, that is $y'(x_0)=2$. Taking into account that $y'(x)=6x$, we conclude that $6x_0=2$, and hence $x_0=\frac{1}{3}$, $y(\frac{1}{3})=3\frac{1}{9}-1=-\frac{2}{3}.$ It follows that the equation of the tangent line is the following:

$y=-\frac{2}{3}+2(x-\frac{1}{3})$ or $y=2x-\frac{4}{3}.$