Arjun Dhaliwal

2022-10-16

Find the equation of the line tangent to the given curve with the given slope. y = 2x^2-4x​, with tangent line of slope 4.

Eliza Beth13

To find the equation of the line tangent to the curve $y=2{x}^{2}-4x$ with a slope of 4, we need to determine the point of tangency and then use the point-slope form of a line.
First, let's find the derivative of the given curve to obtain the slope of the tangent line at any point on the curve. Taking the derivative of $y=2{x}^{2}-4x$, we have:
$\frac{dy}{dx}=\frac{d}{dx}\left(2{x}^{2}-4x\right)=4x-4$
Setting this derivative equal to the given slope of 4, we have:
$4x-4=4$
Simplifying, we find:
$4x=8$
$x=2$
Now, to find the corresponding y-coordinate of the point of tangency, we substitute the value of $x$ into the original curve equation:
$y=2\left(2{\right)}^{2}-4\left(2\right)$
$y=8-8$
$y=0$
Therefore, the point of tangency is $\left(2,0\right)$.
Now that we have the point of tangency, we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by:
$y-{y}_{1}=m\left(x-{x}_{1}\right)$
where $\left({x}_{1},{y}_{1}\right)$ is a point on the line and $m$ is the slope of the line.
Plugging in the values, we have:
$y-0=4\left(x-2\right)$
Simplifying, we find:
$y=4x-8$
Therefore, the equation of the line tangent to the curve $y=2{x}^{2}-4x$ with a slope of 4 is $y=4x-8$.

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